Transcript Lecture 9 Superposition Superposition Theorem (1/2) • In a linear system, the linear responses of linear independent sources can be combined in a linear.

Lecture 9
Superposition
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
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 
a
I12
5
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 
10 V
a
I1
5
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 
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
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
10 
Superposition Example 2 (3/5)
16 
16 A
6
Ixb
16 A
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
10 
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  6 A