Chapter 5
Circuit Theorems
Source Transformations
Replace a voltage source and series resistor by a current and parallel resistor
Figure 5.2-1
(a) A nonideal voltage source.
(b) A nonideal current source.
(c) Circuit B-connected to the nonideal voltage source.
(d) Circuit B connected to the nonideal current source.
Source Transformations
Figure 5.2-2
slope  
Rt 
Figure 5.2-3
voc
isc
voc
isc
 v 
v    oc i  voc
 isc 
or
v   Rt i  voc
Source Transformations
Figure 5.2-4 Thevenin equivalent circuit.
Rt i  v  voc  0
v   Rt i  voc
Figure 5.2-5 Norton equivalent circuit.
isc 
v
i  0
Rt
Figure 5.2-6 Source transformation
v   Rt i  Rt isc
Source Transformations
Example 5.2-1
Example 5.2-2
Find i p , R p , va , vb
ip 
12
 0.002 A
6000
va 
2000
12  3V
2000  6000
vb 
2000 R p
2000  R p
i p  3V
va  vb  3V
R p  6k
va  vb
Source Transformations
Example 5.2-3
Find the relationship between R and i
i
2000
0.001  2
2000  R
2000  R
Source Transformations
Exercise 5.2-1
Exercise 5.2-2
Exercise 5.2-3
Exercise 5.2-4
Superposition
Superposition principle requires that the total effect of several causes acting simultaneously
is equal to the sum of the effects of the individual causes acting one at a time
Example 5.3-1
Express output as a linear combination of inputs
KCL at supernode
v1  v3  vo 
v
 i2  o
40
10
When i2  0 and v3  0
vo1 
10
1
v1  v1
40  10
5
When v1  0 and v3  0
vo 2  10 
40
i2  8i2
40  10
1
1
vo  v1  8i2  v3
5
5
When v1  0 and i2  0
vo 3 
10
 v3    1 v3
40  10
5
Superposition
Example 5.3-2
Find i
Figure 5.3-6
(a) A circuit.
(b) The independent voltage source acting alone.
(c) The independent current source acting alone.
24  3  2i1  3i1
i2  
va
3
i1  3 A
i2  7 
va  3i2
2
va  3i2
 i  i1  i2 
7
i2   A
4
5
A
4
Thevenin’s Theorem
Figure 5.4-2
(a) A circuit partitioned into two parts: circuit A and circuit B.
(b) Replacing circuit A by its Thévenin equivalent circuit.
The open-circuit voltage, voc , the short circuit current, isc , and
the Thevenin resistance, Rt , are related by the equation voc  Rt isc
Figure 5.4-3
Thévenin equivalent circuit involves three parameters:
(a) the open-circuit voltage, voc,
(b) the short-circuit current isc, and
(c) the Thévenin resistance, Rt.
Figure 5.4-4
(a) The Thévenin resistance, Rt
(b) a method for measuring or calculating
the Thévenin resistance, Rt.
Thevenin’s Theorem
Example 5.4-1
Determine Thevenin equivalent circuit
125  voc
v
 2  oc
50
200
voc  20V
125
 2  0  isc
50
isc  0.5 A
Rt  50 // 200  40
Figure 5.4-7
Thevenin’s Theorem
Example 5.4-2
Determine Thevenin equivalent circuit
ia 
KVL
voc
40
voc  96V
0  12  voc  103.5ia 
KVL 5isc  40ia  0
KCL
4.5ia  ib  ia  isc 
KVL
 9 
 12  5isc  10  isc   0
 16 
KCL
4.5ia  it  ia   ib
KVL 40ia  10ib
KVL
isc  1.1294 A
vt  5it  10ib
Rt 
vt
 85
it
Thevenin’s Theorem
Example 5.4-3
Determine
(a) i when R  2
(b) R required to cause i  5 A
(c) R required to cause i  8 A
i
48
R 8
Thevenin’s Theorem
Figure 5.4-16
(a) Circuit under test with
laboratory source vs and resistor R.
(b) Circuit with Thevenin equivalent circuit
replacing test circuit.
v  voc  iRt
Assumes
R 10
Measures two sets of values by experiments
vs  49V : i  0.5 A : v  44V
vs  76V : i  2 A : v  56V
v  voc  iRt
44  voc  0.5Rt
56  voc  2Rt
 Rt  8
voc  40V
Thevenin’s Theorem
Exercise 5.4-1
Exercise 5.4-2
ia 
va  12 2ia  12

6
6
voc  2ia  6V
ia  3 A
KVL
ia  3 A
12  6ia  2ia
3isc  2ia
Rt 
isc 
voc  6

 3
isc  2
2
 3  2 A
3
Norton’s Equivalent Circuit
Source transformation of the Thevenin’s equivalent circuit
Figure 5.5-1
Norton equivalent circuit.
Example 5.5-1
isc  1.125 A
Rt  Rn  32
Norton’s Equivalent Circuit
Example 5.5-2
voc  96V
isc  1.1294 A
Rt  Rn  85
Norton’s Equivalent Circuit
Example 5.5-3
Determine
(a) i when R  2
(b) R required to cause i  5 A
(c) R required to cause i  8 A
v
8R
6  48R
R 8
R 8
Exercise 5.5-1
Rt  8
Mesh #1
3  9i1  6isc
Mesh #2
12isc  6i1  0
 isc  0.25 A
Maximum Power Transfer
Maximum power available from a source be transferred to an output, load resistor RL
Figure 5.6-1
Circuit A with load resistor RL.
2
 vs

p  i 2 RL  
  RL


R

R
L 
 t
Figure 5.6-2
Thevenin equivalent circuit.
vs , Rt
fixed
Find the value of RL that maximize the power
R  RL   2Rt  RL RL  0
dp
 vs2 t
dRL
Rt  RL 4
2
 pmax
vs2 Rt
vs2


2 Rt 2 4 Rt
Figure 5.6-3
Power actually attained as RL varies in relation to Rt.
RL  Rt
dp
0
dRL
Maximum Power Transfer
Figure 5.6-4
Norton’s equivalent circuit.
i  is 
Rt
Rt  RL
is2 Rt2 RL
p  i RL 
Rt  RL 2
2
2
2
dp
2 Rt Rt  RL   2Rt  RL Rt RL
 is
0
dRL
Rt  RL 4
2
 pmax
is2 Rt3
is2 Rt


2 Rt 2 4
RL  Rt
Maximum Power Transfer
Example 5.6-1
Find RL for maximum power delivery
Figure 5.6-6
Thevenin equivalent circuit
connected to RL.
Thevenin equivalent circuit
vt  180 
150
 150V
30  150
Power is maximize when
RL  Rt
150  225W
vs2


4 RL 4  25
Rt  RL  25
2
 pmax
Rt  150 // 30  25
Power at the source
(150V)
i
150
 3A
25  25
ps  vt i  150  3  450W
 pmax  ps  50%
Power at the original source
(180V)
i
180
 3.5 A
(150 // 25)  30
ps  vi  180  3.5  630W
 pmax  ps  35.7%
Maximum Power Transfer – with dependent sources
Example 5.6-2 Find RL for maximum power delivery
6  2vab  (6  4)i
vab  voc  4i
2vab  0 vab  0
6  6isc
isc  1A
Rt 
voc 12V

 12
isc
1A
Rt  RL  12
 pmax
voc2

 3W
4 RL
i  3 A, voc  12V
Maximum Power Transfer
Exercise 5.6-1
Rt  4
Rt  RL  4
vt  12V
 pmax
vt2

 9W
4 RL
MatLab
Checking Thevenin Equivalent Circuit
Example 5.8-1
Find
vOC
and
isc
vOC  vR
when
vSC  iR 
vR
when R  5
R
R to a very large value
R
R  20
to a very small value
vOC
12.6 10 3
isc 
 12.6 A
110 3
Figure 5.8-6
simulation results for R  20M
Figure 5.8-7
simulation results for R  1m
Checking Thevenin Equivalent Circuit
Example 5.9-1
v
Experiment #1
Experiment #2
R  2k
R  5k
v  1.87V
v  3.00V
 1.87 
 3.00 
By calculation when R  10k
Experiment #3
R  10k
v  3.75V
2,000
voc
2,000  Rt
5,000
voc
5,000  Rt
v
R
voc
R  Rt
voc  5V
Rt  3,333
10,000
(5)  3.75V
10,000  3,333
v
R
(5)
R  3,333
Straingage Bridge
Find amplifier gain, b
Voltage source : 50mV
Nominal resistance without strain : R 120
Range of resistance change :  2  R  2
Figure 5.10-1
Design problem involving a Strain gauge bridge.
Figure 5.10-2
Output voltage :  10V  vo  10V
Amplifier gain : vo  5
Figure 5.10-3
V
 R

Straingage Bridge
Figure 5.10-3
Figure 5.10-2
i1 
50mV
50mV

R  R   R  R  2R
vi  vt 
100k
 0.9988vt  0.4162 10 3 R
100k  Rt
i2 
50mV
50mV

R  R   R  R  2R
KVL
vo  50(0)  bvi
50mV
vt  R  R i1  R  R i2  2R 
2R
R
R

50mV 
50mV  0.4167 10 3 R
R
120

Rt 
R  R R  R   R  R R  R 
R  R   R  R  R  R   R  R 
R 2  R 2
 2
R
2R
since R  R





vo  bvi  b 0.4162 103 R


b 0.4162 103  5 since vo  5R
b  12.013
Substituting b  12.013


vo  12.013 0.4162 103 R  4.9998R
Homework #5
Problems
P5.2-2 / P5.2-8 / P5.3-8 / P5.3-14 /
P5.4-4 / P5.4-13 / P5.5-3 / P5.5-9 /
P5.6-1 / P5.6-8