Transcript Electrical Engineering 105

Electrical Systems 100
Lecture 2
Dr Kelvin Tan
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Contents
•Series-Parallel Circuits
• Wye-Delta Conversion
• Ladder Networks
• Current Sources
• Source Conversion
• Current Sources in Parallel (Series?)
• Mesh Analysis
• Nodal Analysis
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Series-Parallel Circuits
Series-parallel circuits are networks where there are
both series and parallel elements
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Reduce and Return Approach
Step 1: Take an overall mental look at the problem
Step 2: Examine each section of the network independently before
tying them together
Step 3: Redraw the network as often as you will need to arrive at
reduced branches. Maintain the original unknown quantities to be
found for clarity where applicable
Step 4: Take the trip back to the original network to find detailed
solution
(Some time it may help to draw branche/s as blocks and work out
blockwise)
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An Example of Reduced and Return Approach
Finding V4?
IS
IS
R1
IS 
E
RT
IS
RT  R'T  R1
R'T
V2
R'T  ( R3  R4 ) // R2
V2  I S  RT
V2
V4  V2 
R4
R3  R4
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An Example of Reduced and Return Approach
2kΩ
12kΩ //6kΩ
54V
IS 
E
2k  12kΩ //6kΩ 
6kΩ
 IS
12kΩ  6kΩ
12kΩ
I2 
 IS
12kΩ  6kΩ
I1 
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Wye-Delta Conversion
Often we encounter a different kind of network which appears to be
not in series or parallel in relation to the rest of the network. Under
these circumstances it is necessary to convert this portion of circuit
from one form to the other to find appropriate branch connection
which then appears clearly in series or parallel with rest of the
network.
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Y- Conversion
The purpose is then to be able to convert Y to  or  to Y.
Ra c  R1  R3  R B //( R A  RC )
Ra c
R B ( R A  RC )
 R1  R3 
R B  ( R A  RC )
RB R A

R B  R A  RC
R B RC

R B  R A  RC
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Ra  c
R B ( R A  RC )
R B RC
RB R A
 R1  R3 


R B  ( R A  RC )
R A  R B  RC
R A  R B  RC
Ra b  R1  R2 
RC R A
RC R B
Rc( R A  R B )


RC  ( R A  R B )
R A  R B  RC
R A  R B  RC
Rb  c  R2  R3 
R A ( R B  RC )
R A RC
R A RB


R A  ( R B  RC )
R A  R B  RC
R A  R B  RC
-Y Conversion
RB RC
R1 
R A  RB  RC
R A RC
R2 
R A  RB  RC
R A RB
R3 
R A  RB  RC
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Y- Conversion
Similarly, for converting Wye quantities to Delta are given as:
R1 R2  R2 R3  R1 R3
RA 
R1
R1 R2  R2 R3  R1 R3
RB 
R2
R1 R2  R2 R3  R1 R3
RC 
R3
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-Y /Y- Conversion
If all resistors in the  or Y are the same (RA = RB = RC):
From -Y eq:
RA RB
RA2
R3 

RA  RB  RC 3RA
RA

 R2  R1
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This shows that for a Y of three equal resistors the value of each
resistor equivalent  is 3 times the Y resistor.
RΔ
For Δ  Y : R Y 
3
For Y  Δ : R Δ  3R Y
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Ladder Networks
A Ladder Network is one where a series-parallel section of a network occur
repeatedly within the network. An example of such network is a Low Pass Filter
circuit. Figure below shows a three section Ladder Network.
To solve a ladder network follow the steps:
• Calculate the total resistance
• Calculate the source current or total current drawn from source
• Work back through the ladder until desired current or voltage is obtained
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Ladder Networks
Combining parallel and series elements to reduce the circuit we get :
RT  5Ω  3Ω  8Ω
IS 
E
240V

 30 A
RT
8Ω
W orkingbackwards, I1  I S
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Ladder Networks
Using current divider I6 can be found.
IS
 15 A
2
I 2  15 A
I S  30 A,  I 3 
(Current divider)
Finally,
6
6
 I 3  I 3  10 A
(6  3)
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and V6  I 6 R 6  10A 2Ω  20 V
I6 
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Current Sources
A battery supplies fixed voltage and the source current may vary
according to load. Similarly, a current source is one where it
supplies constant current to the branch where it is connected and
the voltage and polarity of voltage across it may vary according to
the network condition.
VS  E  12 V
12V
I2 
 3 A, ApplyingKCL
4
I1  I  I 2  7  3  4 A
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Source Conversion
A Voltage source can be converted to a current source and vice
versa. In reality, Voltage sources has an internal resistance Rs and
current sources has a shunt resistance Rsh. In ideal cases, Rs equal
to 0 and Rsh equal to .
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Source Conversion
For us to be able to convert sources, the voltage source must have a series
resistance and current source must have some shunt resistance.
Eg.
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Current Sources in Parallel
If two or more sources are in parallel, then the equivalent
source is obtained by summing the currents with direction
taken into account and the new shunt resistance is the
parallel combination of all the individual resistances.
Note that current sources can not be connected in series!
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Method of Circuit Analysis
(Mesh Analysis and Nodal Analysis)
Mesh is a closed loop which does not contain any other loops
within it. In most circumstances, a mesh will contain one or more
voltage sources and one or more type of circuit elements. In dc
circuit theory, these elements are limited to resistances only in
steady state analysis. Mesh analysis determines the mesh or loop
currents in the circuit.
By solving i1 and i2
I1 = i 1
I2 = i 2
I3 = i1-i2
i1
i2
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Mesh Analysis
Steps to determine Mesh Currents:
• Identify the “n” number meshes in the circuit
• Assign mesh currents, i1, i2 , i3......in1,in
in clockwise directions
• Apply KVL to each of the “n” meshes. Use Ohm’s law to
express the voltages in terms of the mesh currents. Take
appropriate voltage drop polarity (+ve clockwise and –ve
anticlockwise) into consideration in writing these equations.
• Solve the “n” simultaneous equations to get the “n” mesh
currents.
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Mesh Analysis-An Example
Find I1, I2 and I3.
i2
i1
(Loop i1) 15  5i1  10(i1  i 2 )  10  0
5  15i1  10i2  0
3i1  2i2  1
(Loop i2) 10  10(i2  i1 )  6i2  4i2  0
Method 1:Using method of substitution
6i2  3  2i2  1
i 2  1 A and i1  1 A.
10  10i1  20i2  0
2i2  i1  1
i1  2i2  1
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Mesh Analysis-An Example
Method 2: Using Cramer’s Rule (Also known as Format Approach):
From previous 2 loops : 3i  2i  1
1
2
 i1  2i2  1
 3  2  i1  1
 1 2  i   1

 2   
We obtain the determinant Δ as
3 -2
Δ
 62  4
-1 2
Δ1 
Δ2 
1 2
 22  4
1 2
3
1
1 1
Thus:
Δ1
i1 
 1A
Δ
Δ
i2  2  1A
Δ
 3 1  4
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Mesh Analysis-Super Mesh
Sometime, there may be a current source in one of the mesh or a
current source in common between two meshes.
• If the current source involves only one of the meshes, then the
analysis is easier as we have 1 less equation to solve as the current is
defined by the current source in one of the mesh already.
• If however, the current source is common between two meshes,
then we need to form a super mesh. This is necessary as we need to
apply KVL in solving mesh equations and we do not know the
voltage across the current source in advance.
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Mesh Analysis-Super Mesh
Case 1:
i1
In this example,
i2
i2  5 A.
Applying KVL in Mesh 1
- 10  4i1  6(i1  i2 )  0
i1  2 A
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Mesh Analysis-Super Mesh
Case 2:
i1
i1
i2
i 2  i1  6 A
i2
super-mesh
we create a super-mesh by excluding the current source and any
element connected in series with it as shown.
In super-mesh
 20  6i1  10i2  4i2  0
6i1  14i2  20
i1  3.2 A
i 2  2.8 A
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Nodal Analysis
A Node is defined as the junction of two or more branches. In a “n”
node circuit, 1 node is taken as the reference (usually the ground is
taken as reference) and we need to solve node voltages using KCL.
By solving node v1 and v2
We can solve :
I1 
V1
R1
I2 
V1  V2
R2
I3 
V2
R3
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Nodal Analysis
Steps to determine Nodal voltages:
• Select a node as reference node
• Assign voltages to the remaining nodes
• Apply KCL to each of the n-1 non-reference nodes. Use Ohm’s law
to express currents in terms of node voltages.
• Solve the n-1 simultaneous equations to get the unknown node
voltages.
v1 , v2 ,....vn1
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Nodal Analysis
Applying KCL to each nodes
Example in node v1
V1  0 V1  V2

 I 2  I1  0
R1
R2
node v2
V2  0 V2  V1

 I2  0
R3
R2
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Nodal Analysis- An Example
At node 1, applying KCL,
v1  0
v  v2
 1
5  0
2
4
v1
v2
20  2v1  v1  v 2
3v1  v 2  20
At node 2, applying KCL,
v 2  0 v 2  v1

 5  10  0
6
4
v1
v2
- 3v1  5v 2  60
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Nodal Analysis- An Example
Using Cramer’s Rule or Format Approach:
 3  1  v1  20
 3 5  v   60

 2   
3 1

 15  3  12
3 5
20  1
1 60 5 100 60
v1 


 13.33V


12
3 20
 2  3 60 180 60
v2 


 20V


12
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Nodal Analysis-Supernode
Sometime, there may be a voltage source connected between a
reference and nonreference node. If this is the case, the voltage of
the nonreference node is simply set equal to the voltage source and
we have 1 less equation to solve.
If however, the voltage source is common between two or more
unknown nodes, then we need to form a super node. This is
necessary as we need to apply KCL in solving node equations and
we do not know the current through the voltage source in advance.
A Supernode is formed by enclosing the voltage source between
two nonreference nodes and any branches in parallel with it.
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Nodal Analysis-Supernode
Consider the following circuit. Nodes 2 and 3 form a supernode. At
Supernode we get,
v1
v2
v3
v1
v2
v3
At Supernode, applying KCL,
v3  0
v 3  v1
v 2  v1
v2  0



0
2
8
6
4
v1  10V
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Nodal Analysis-Supernode
Applying KVL to supernode we get,
 v2  5  v3  0
or, v2  v3  5
v
v
2
3
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Nodal Analysis-Supernode
We note the following properties of a supernode,
• The voltage source inside the supernode provides a constraint
equation to solve for the node voltages
• A supernode do not have a voltage of its own
• A supernode requires the application of both the KCL and KVL
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