Slide 1

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 2

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 3

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 4

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 5

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 6

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 7

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 8

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 9

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 10

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 11

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 12

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 13

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 14

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 15

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 16

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 17

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 18

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 19

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 20

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 21

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 22

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 23

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 24

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 25

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 26

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 27

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28


Slide 28

Chapter 17 – Methods of Analysis & Sel Topics
Lecture 23
by Moeen Ghiyas

31/10/2015

1

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

2

 Independent Sources
 The term independent specifies that the magnitude of the

source is independent of the network to which it is applied and
that the source displays its terminal characteristics even if
completely isolated.

 Dependent (Controlled) Sources
 A dependent or controlled source is one whose magnitude is

determined (or controlled) by a current or voltage of the system in
which it appears
Old
Symbols

New
Symbols

 Dependent (Controlled) Sources
 Unlike with the independent source, isolation such that V or I = 0

will result in short-circuit or open-circuit equivalent as indicated

 Source conversion can be accomplished in much the same

manner as for dc circuits, except now we shall be dealing
with phasors and impedances instead of just real numbers
and resistors

 EXAMPLE - Convert the voltage source of fig to a current source

 Solution:

 EXAMPLE - Convert the current

source of fig to a voltage source
 Solution:

 For dependent sources, the direct conversion can be

applied if the controlling variable (V or I) is not determined
by a portion of the network to which the conversion is to be
applied
 Conversions of the other kind, where V and I are controlled

by a portion of the network to be converted, are covered in
chapter 18 (Not part of syllabus for 2nd semester)

 EXAMPLE - Convert the voltage source of fig to a current source
 Solution:

 Steps

1.

Assign a distinct current
direction to each independent,
closed loop

2.

Indicate the polarities within
each loop for each resistor
(impedance) . Note that the

polarities be placed within each
loop. Thus 4 Ω resistor have
two sets of polarities across it.

 Steps
3.

Apply Kirchhoff’s voltage law around each closed loop
 the total current through the element (impedance) is the assumed

current of the loop plus the assumed currents of the other loops
passing through in the same direction, minus the assumed currents
through in the opposite direction
 The polarity of a voltage source is unaffected by the direction of the

assigned loop currents
4.

Solve the resulting simultaneous linear equations for the
assumed loop currents

 The general approach to mesh includes the same sequence of

steps as for dc except to substitute impedance for resistance and

admittance for conductance in the general procedure with minor
additional changes as mentioned below:
 Independent Voltage Sources

 Same as dc analysis with impedances and admittance values
 Dependent Voltage Sources
 Step 3 is modified: Treat each dependent source like an

independent source when KVL is applied. However, once the
equation is written, substitute the equation for the controlling
quantity (i.e an additional eqn is generated for controlling qty)

 Independent Current Sources
 Treat each current source as an open circuit (recall supermesh

concept), and write mesh equations for remaining paths.
 Then relate the chosen mesh currents to the independent

sources to ensure that the unknowns of the final equations are
limited simply to the mesh currents
 Dependent Current Sources
 The procedure is same as for independent current sources (i.e.

supermesh concept), except now the dependent sources have
to be defined in terms of the chosen mesh currents so that the
final equations have only mesh currents as the unknown qtys.

 EXAMPLE - Using the general approach to mesh analysis, find

the current I1 in Fig
 Solution

 The network is redrawn

 Apply KVL,

Loop 1:
+E1 – I1Z1 – I1Z2 + I2Z2 = 0
I1Z1 + I1Z2 - I2Z2 = E1
I1(Z1 + Z2) – I2Z2 = E1

----- A

Loop 2:
– E2 – I2Z2 + I1Z2 – I2Z3 = 0
– I1Z2 + I2Z2 + I2Z3 = – E2
– I1Z2 + I2(Z2 + Z3) = – E2 ------ B
 Solve by determinants

 Solve by determinants and then substitute values

I1(Z1 + Z2) – I2Z2 = E1
– I1Z2 + I2(Z2 + Z3) = – E2

 EXAMPLE - Write the mesh currents for the network of fig having

a dependent voltage source.
 Solution:

 EXAMPLE - Write the mesh currents for the network of fig having

a independent current source.
 Solution:
 Apply supermesh concept

 EXAMPLE - Write the mesh currents for the network of fig

having a dependent current source.
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Problem # 9 – Using mesh analysis, determine the current IL (in

terms of V) for the network of fig
 Solution:

 Solving by determinants

 Bridge networks with reactive components & ac voltage or current
 Maxwell’s Bridge (V source & RC parallel)

 From dc we remember, for IZ5 = 0,

the following condition must be met

or

 Hay Bridge (I source & RC series)

 From dc we remember, for VZ5 = 0,

the following condition must be met

 or

 or

 Hay bridge – when Z5 is replaced by a sensitive galvanometer is

used for measuring the resistance and inductance of coils in
which the resistance is a small fraction of the reactance XL.
 Maxwell bridge – when Z5 is replaced by a sensitive galvanometer

is used for inductance measurements when the resistance of the
coil is large enough not to require a Hay bridge.

 Another popular bridge is the capacitance comparison bridge

of fig. An unknown capacitance and its associated resistance

can be determined using this bridge.

 Independent vs Dependent Sources
 Source Conversions
 Mesh Analysis
 Bridge Networks

31/10/2015

28