Download E E 2315 Circuits I Lecture 7

E E 2315
Lecture 07 - 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
VTh
I
RTh  RL
RL
b
dPL
PL is max if
0
dRL
and
PL  V
2
Th
RL
 RTh  RL 
2
Maximum Power Demonstrated (2/2)
RTh
VTh
a
I
dPL
0
dRL
 RL  RTh
RL
b
and
• Maximum Power Transfer Theorem applied
when matching loads to output resistances
of amplifiers
• Efficiency is 50% at maximum power
transfer
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 
10 V
a
I1
5
I2
5A
b
Find I1, I2 and Vab by superposition
Superposition Example 1 (2/4)
Step 1: Omit current source.
20 
10 V
a
I11
5
I21
5A
b
By Ohm’s law and the
voltage divider rule:
I11  I 21  0.4 A
Superposition Example 1 (3/4)
Step 2: Omit voltage source.
20 
a
I12
5
I22
b
By the current divider
rule and Ohm’s law :
5A
I12  1.0 A
Vab 2  20V
Superposition Example 1 (4/4)
20 
10 V
a
I1
5
I2
5A
b
Combining steps
1 & 2, we get:
I1  I11  I12  0.6 A
Vab  Vab1  Vab 2  22V
Superposition Example 2 (1/5)
16 
16 A
6
Ix
64 V
16 A
Find Ix by superposition
10 
Superposition Example 2 (2/5)
16 
16 A
6
Ixa
16 A
10 
Activate only the 16 A Current source at the left. Then
use Current Divider Rule:
10  16
I xa 
16 A  13 A
6  10  16
Superposition Example 2 (3/5)
16 
16 A
6
Ixb
16 A
10 
Activate only the 16 A Current source at the right. Then
use Current Divider Rule:
10
I xb  
16 A  5 A
6  10  16
Superposition Example 2 (4/5)
16 
16 A
6
Ixc
64 V
16 A
10 
Activate only the 64 V voltage source at the bottom. Then
use Ohm’s Law:
64V
I xc  
 2 A
6  10  16
Superposition Example 2 (5/5)
16 
16 A
6
Ix
64 V
16 A
10 
Sum the partial currents due to each of the sources:
I x  I xa  I xb  I xc  13 A  5 A  2 A 
Superposition Example 3 (1/4)
4
12 V
a
ix
2
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)
4
a
ix
2
1.5ix
30 V
b
Vab1
Vab1  30
0
 1.5ix 
4
2
1
 1  1  9
15  Vab1    1.5      Vab1
 4  2 8
4
40
Vab1  V
3
Superposition Example 3 (3/4)
4
12 V
a
ix
2
1.5ix
b
Vab 2
Vab 2  12
0
 1.5ix 
2
4
 1 1.5 1  9
7.5  Vab 2   
   Vab 2
2 4 4 8
Vab 2
20

3
Superposition Example 3 (4/4)
4
12 V
a
ix
2
1.5ix
b
30 V
Vab  Vab1  Vab 2
40
20
Combining the solutions: Vab 
V V
3
3
Vab 