E E 2415
Lecture 04 - Network
Theorems
Maximum Power Transfer Theorem
• Given a Thévenin equivalent circuit of a
network, the maximum power that can be
transferred to an external load resistor is
obtained if the load resistor equals the
Thévenin resistance of the network. Any
load resistor smaller or larger than the
Thévenin resistance will receive less power.
Maximum Power Demonstrated (1/2)
RTh
VTh
a
I
RL
b
and
PL  I RL
RL
2
PL  V
PL is max if
dPL
0
dRL
VTh
I
RTh  RL
2
Th
 RTh  RL 
2
R R
dPL
2
 VTh
4
dRL
 RTh  RL 
2
Th
2
L
Maximum Power Demonstrated (2/2)
RTh
dPL
 0  RL  RTh
a
dRL
VTh
I
RL
b
and
PLmax
2
Th
V

4 RTh
• Maximum Power Transfer Theorem applied
when matching loads to output resistances of
amplifiers
• Efficiency is 50% at maximum power transfer
Max Power Plot
Vth = 100 V
Rth = 10 W
Superposition Theorem (1/2)
• In a linear system, the linear responses of
linear independent sources can be combined
in a linear manner.
• This allows us to solve circuits with one
independent source at a time and then
combine the solutions.
– If an independent voltage source is not present it
is replaced by a short circuit.
– If an independent current source is not present it
is replaced by an open circuit.
Superposition Theorem (2/2)
• If dependent sources exist, they must remain
in the circuit for each solution.
• Nonlinear responses such as power cannot be
found directly by superposition
• Only voltages and currents can be found by
superposition
Superposition Example 1 (1/4)
20 W
10 V
a
I1
5W
I2
5A
b
Find I1, I2 and Vab by superposition
Superposition Example 1 (2/4)
Step 1: Omit current source.
20 W
10 V
a
I11
5W
I21
b
5A
By Ohm’s law and the
I11  I 21  0.4 A
voltage divider rule:
Vab1  2.0V
Superposition Example 1 (3/4)
Step 2: Omit voltage source.
20 W
a
I12
5W
I22
b
By the current divider
rule and Ohm’s law :
5A
I12  1.0 A
I 22  4.0 A
Vab 2  20V
Superposition Example 1 (4/4)
20 W
10 V
a
I1
5W
I2
5A
b
Combining steps
1 & 2, we get:
I1  I11  I12  0.6 A
I 2  I 21  I 22  4.4 A
Vab  Vab1  Vab 2  22V
Superposition Example 2 (1/5)
16 W
16 A
6W
Ix
64 V
16 A
Find Ix by superposition
10 W
Superposition Example 2 (2/5)
16 W
16 A
6W
Ixa
16 A
10 W
Activate only the 16 A Current source at the left. Then
use Current Divider Rule:
10W  16W
I xa 
16 A  13 A
6W  10W  16W
Superposition Example 2 (3/5)
16 W
16 A
6W
Ixb
16 A
10 W
Activate only the 16 A Current source at the right. Then
use Current Divider Rule:
10W
I xb  
16 A  5 A
6W  10W  16W
Superposition Example 2 (4/5)
16 W
16 A
6W
Ixc
64 V
16 A
10 W
Activate only the 64 V voltage source at the
bottom. Then use Ohm’s Law:
64V
I xc  
 2 A
6W  10W  16W
Superposition Example 2 (5/5)
16 W
16 A
6W
Ix
64 V
16 A
10 W
Sum the partial currents due to each of the sources:
I x  I xa  I xb  I xc  13 A  5 A  2 A  6 A
Superposition Example 3 (1/4)
4W
12 V
a
ix
2W
1.5ix
30 V
b
• Solve for Vab by Superposition method
• Dependent source must remain in circuit for
both steps
Superposition Example 3 (2/4)
4W
ix
a
2W
1.5ix
30 V
b
Vab1
Vab1
Vab1  30
ix  
0
 1.5ix 
4
4
2
1
40
 1  1  9
15  Vab1    1.5      Vab1 Vab1  V
3
 4  2 8
4
Superposition Example 3 (3/4)
4W
12 V
a
ix
2W
1.5ix
b
12  Vab 2
Vab 2
Vab 2  12
ix 
0
 1.5ix 
4
2
4
20
 1 1.5 1  9
7.5  Vab 2   
   Vab 2 Vab 2 
3
2 4 4 8
Superposition Example 3 (4/4)
4W
12 V
a
ix
2W
1.5ix
b
30 V
Vab  Vab1  Vab 2
40
20
Combining the solutions: Vab  V  V
3
3
Vab  20V