Transcript Basic Electronics Ninth Edition Grob Schultz ©2002 The McGraw-Hill Companies Basic Electronics Ninth Edition CHAPTER Kirchhoff’s Laws ©2003 The McGraw-Hill Companies Topics Covered in Chapter 9  Kirchhoff’s Current Law  Kirchhoff’s.

Basic Electronics
Ninth Edition
Grob
Schultz
©2002 The McGraw-Hill Companies
Basic Electronics
Ninth Edition
CHAPTER
9
Kirchhoff’s Laws
©2003 The McGraw-Hill Companies
Topics Covered in Chapter 9
 Kirchhoff’s Current Law
 Kirchhoff’s Voltage Law
 Branch Current Analysis
 Node Voltage Analysis
 Mesh Current Analysis
Kirchhoff’s Current Law
The sum of currents entering any
point in a circuit is equal to the sum of
currents leaving that point.
Example:
I1
I3
I2
I1 + I 3 = I2
or
I1 - I2 + I 3 = 0
Kirchhoff’s Voltage Law
The algebraic sum of the voltage rises and
IR voltage drops in any closed path must
total zero.
Example:
VT - VR1 - VR2 = 0
or
VT = VR1 + VR2
VR1
VT
VR2
Branch Currents
• A loop is a closed path.
• This approach uses the algebraic
equations for the voltage around the
loops of a circuit to determine the
branch currents.
 Use the IR drops and KVL to write the
loop equations.
Branch Current Method
R1
V1
R2
I1
I2
I3
R3
V2
VR1 = I1R1 VR2 = I2R2 VR3 = I3R3 also VR = (I1+I2)R3
3
Loop equations:
V1 – I1R1 – (I1+I2) R3 = 0
V2 – I2R2 – (I1+I2) R3 = 0
Branch Current Method Applied
100 W
R1
15 V
V1
20 W
R2
I1
I2
R3
I3
10 W
13 V
V2
V1 – I1R1 – (I1+I2) R3 = 0
V2 – I2R2 – (I1+I2) R3 = 0
15 – 100I1 – 10(I1+I2) = 0
13 – 20I2 – 10(I1+I2) = 0
15 – 100I1 – 10I1 – 10I2 = 0
13 – 20I2 – 10I1 – 10I2 = 0
15 – 110I1 – 10I2 = 0
13 – 10I1 – 30I2 = 0
Branch Current Method Applied
100 W
R1
15 V
V1
20 W
R2
I1
I2
R3
I3
10 W
13 V
V2
-3[ 15 – 110I1 – 10I2 = 0 ]
13 – 10I1 – 30I2 = 0
-45 + 330I1 + 30I2 = 0
}Add the equations
13 – 10I1 – 30I2 = 0
-32 + 320I1 = 0
320I1 = 32
I1 = 0.1 A
Branch Current Method Applied
100 W
R1
15 V
V1
20 W
R2
I1
I2
R3
I3
10 W
15 – 110I1 – 10I2 = 0
13 V
V2
13 – 10I1 – 30I2 = 0
Substitute: 13 – 1 – 30I2 = 0
12 – 30I2 = 0
I2 = 0.4 A
I3 = 0.1 + 0.4 = 0.5 A
I1 = 0.1 A
Branch Current Method Applied
15 V
0.5 A
–
1
20 W
+
–
+
0.4 A
10 W
+
100 W
+
–
0.1 A
–
+
–
13 V
2
KVL check (loop 1): 15 V – 5 V – 10 V = 0
KVL check (loop 2): 13 V – 5 V – 8 V = 0
KVL check (outside loop): 15 V – 13 V + 8 V – 10 V = 0
(The 13 V battery is a drop and V100 W is a rise.)
Node Voltage Analysis
• A principal node is a point where
currents divide or combine, other than
ground.
• The method of node voltage analysis uses
algebraic equations for the node currents
to determine each node voltage.
Use KCL to determine node currents
Use Ohm’s Law to calculate the voltages
Node Voltage Method
R1
V1
R2
N
I1
I2
R3
I3
At node N: I1 + I2 = I3
or
VR1
R1
+
VR2
R2
=
VN
R3
V2
Node Voltage Method Applied
R1
15 V
V1
R2
N
100 W
20 W
10 W
VR1
R1
VR1
100
+
+
R3
VR2
R2
VR2
20
=
=
VN
R3
VN
10
13 V
V2
Node Voltage Method Applied
R1
100 W
15 V
V1
VR1
100
+
VR2
20
R2
N
10 W
=
20 W
R3
13 V
V2
VN
10
VR1 + VN = 15 or VR1 = 15 – VN
VR2 + VN = 13 or VR2 = 13 – VN
VN
15 – VN
13 – VN
Substitute:
=
+
100
20
10
Node Voltage Method Applied
5V
N
10 V
15 V
V1
100 W
10 W
8V
20 W
R3
13 V
V2
VN
15 – VN
13 – VN
=
+
100
20
10
VN = 5
This agrees with the solution found using
the branch method.
Mesh Current Analysis
• A mesh is the simplest possible loop.
• Mesh currents flow around each
mesh without branching.
• IR drops and KVL are used for
determining mesh currents.
Mesh Current Method
R1
V1
R2
V2
R3
A
B
A clockwise assumption is standard. Any drop in
a mesh produced by its own mesh current is
considered positive.
Mesh A: R1IA + R3IA – R3IB = – V1
Mesh B: R2IB + R3IB – R3IA = V2
Mesh Current Method Applied
R1
15 V
V1
100 W
10 W
A
R2
20 W
R3
13 V
V2
B
The mesh drops are written collectively here:
Mesh A: 110IA – 10IB = – 15
Mesh B: – 10IA + 30IB = 13
Mesh Current Method Applied
Mesh A: 110IA – 10IB = – 15
R1
15 V
V1
100 W
10 W
A
Mesh B: – 10IA + 30IB = 13
R2
20 W
13 V
V2
R3
B
Multiply equation A by 3 to allow
cancellation of an unknown.
330IA – 30IB = – 45
Add the equations. – 10IA + 30IB =
320IA
13
= – 32
IA = – 0.1 A
Mesh Current Method Applied
Mesh A: 110IA – 10IB = – 15
R1
15 V
V1
100 W
10 W
A 0.5 A
Mesh B: – 10IA + 30IB = 13
R2
20 W
R3
B
13 V
V2
IA = – 0.1 A
110(– 0.1) – 10IB = – 15
IB = 0.4 A
The assumed direction of IA was wrong because
it solved as negative. The magnitude is correct.
Reverse the flow arrow in mesh A.
Substitute.
Both mesh currents flow up through R3. Thus, they add:
IR3 = 0.1 + 0.4 = 0.5 A