Questions and Problems
1. What does a negative response in superposition imply?
A negative response implies that the original assumed direction of current or voltage is
wrong, therefore should be reversed
2. How many responses are obtained from an N number of independent sources present in a
given circuit?
There are no responses obtained simply by having the number of independent sources
present. Responses in a circuit can be voltage across a resistor, or current in a branch. These
parameters cannot be affected the number of sources in a given circuit.
3. Is it possible to eliminate dependent sources on superposition?
No, it is not possible to eliminate independent sources.
4. What are the possible limitations of the superposition theorem?
As what the definition of the superposition theorem implies, it cannot be used to circuits
that are nonlinear. Nonlinear circuits are those that have their parameter values change. This
means that resistance, capacitance, or other parameters are not constant.
5. Determine the voltage Vx using superposition shown in the figure below.
Vx
Considering 10V ON and other sources OFF:
1.6
V 'x =10 ( 2+1.6 )V =4.44 V
'
x
Considering 15 V ON and other sources OFF:
V 'x' =15 ( 1+81 )V =1.67 V
''
x
Considering 10 A ON and other sources OFF:
( 5+5 4 )
I 'x' =10
2
I =5.56 A I =5.56 (
2+2 )
'' '' ''
x x I =2.78 A
x
V 'x' ' =2.78 ( 2 )
V 'x' ' =5.56 V
Adding all the voltages caused by separate sources acting alone
V x =V +V 'x' +V 'x' '
'
x
V x =4.44 V +1.67 V +5.56 V
V x =11.67
6. Determine the voltage across 5ohm resistance using superposition shown in the figure
below.
Considering 12V ON and 7 A OFF:
2.5454
V 5=12 ( 2+2.5454 )
V 5=6.72V
V ' 5=6.72 ( 5+5 2 )
V ' 5=4.8 V
Considering 7A ON and 12V OFF:
Using Nodal Analysis:
Node 1: (0.5+ 0.2)V 1+ 0.2V 2 – 0.5 V 3=−7 Node 2:
0.2 V 1+(0.2+0.25+ 0.5)V 2−(0.5+0.25)V 3=0
Node 3: (0.5+ 0.25+ 0.5)V 3−0.5V 1 – (0.25+0.5) V 2=7
V 1=−8.4 V ; V (2)=0
V 3=2.24 V
V '5' =−8.4−0
V 5=V '(5) +V '(5' )
V 5=4.8−8.4
V 5=−3.6V
7. Determine the voltage across 3mho using the superposition shown in the circuit below.
Considering 6A ON and other sources OFF:
−6+ 2 ( V 1−V 2) + 5 ( V 1−V 2 )=0
Equation 1:
7 V 1−7 V 2=6−( 5+2) V 1 +13 V 2=0
Equation 2:
−7 V 1 +13 V 2=0
V '1=1.857 V ; V '2=1 V
Considering 12A ON and other sources OFF:
Node 1:
7 V 1−7 V 2=0
Node 2:
−7 V 1 +3 V 2=12
V 1=2 V ; V 2=2 V
Considering 10V ON and other sources OFF:
Node 1:
7 V 1−5 V 2−2 V 3=0
Node 2:
−7 V 1 +8 V 2−5 V 3=0
Node 3:
V 2−V 3 =10
V 1=2.143 V ; V 2=5 V ;V 3=−5 V
Considering -3A ON and other sources OFF:
Mesh 1:
0.33 I 1+0.33 I 1=0
I 1=0
V (3 ℧ ) =(0.33) I 1
V (3 ℧ ) =0 V
Total voltage at right:
V 3 ℧ =V '2 +V '2' +V '2' ' +V 3 ℧
V 3 ℧ =1+2+5+0
V 3 ℧ =8 V
Total voltage at left:
V 3 ℧ =V '2 +V '2' −V '2' ' + V 3 ℧
V 3 ℧ =1+2+ (−5 ) +0
V 3 ℧ =−2 V
