Download Example 1: Consider the circuit shown in Figure 1

Superposition
Introduction
Each circuit in this problem set has two inputs and one output. The inputs to these circuits are the
voltages of independent voltage sources and the currents of independent current sources. The
output of each circuit is the voltage or current measure by the meter.
These circuits can be analyzed using superposition. Superposition says that the output caused by
two inputs working together can be determined by finding the outputs due to each input working
separately and then adding those separate outputs.
Superposition is discussed in Section 5.4 of Introduction to Electric Circuits by R.C. Dorf and
J.A Svoboda.
Worked Examples
Example 1:
Consider the circuit shown in Figure 1. Find the value of the current source current, Ia.
Figure 1 The circuit considered in Example 1.
Solution: The inputs to this circuit are the voltage of the independent voltage source and the
current of the independent current source. The response of this circuit is the current measured by
the ammeter. The two inputs work together to produce a response that has a value of 3 A. This
circuit can be analyzed using superposition to separate the part of the response caused by the
voltage source from the part of the response caused by the current source.
Figure 2 shows the circuit from Figure 1 after replacing the ammeter by an equivalent
short circuit and labeling the current measured by the ammeter.
Figure 3a shows the circuit used to determine, i1, the part of the response caused by the
voltage source. The current source current is set to zero to determine the part of the response
1
caused by the voltage source. Consequently, the current source is replaced by an open circuit, i.e.
a zero current source, in Figure 3a.
Figure 3b shows the circuit used to determine, i2, the part of the response caused by the
current source. The voltage source voltage is set to zero to determine the part of the response
caused by the current source. Consequently, the voltage source is replaced by a short circuit, i.e.
a zero voltage source, in Figure 3b.
Figure 2 The circuit from Figure 1 after replacing the ammeter by a short circuit.
(a)
(b)
Figure 3 The circuit from Figure 2 after using superposition. (a) The response to the voltage
source acting alone. (b) The response to the current source acting alone.
Apply KVL to the circuit in Figure 3a to get
2 i1 + 2 i1 − 4 = 0 ⇒ i1 = 1 A
Use current division in Figure 3b to get
i2 =
2
1
Ia = Ia
2+2
2
2
The response to both sources working together is equal to the sum of the responses to the
individual sources working separately. Consequently,
3 = i1 + i2 = 1 +
1
Ia
2
⇒ Ia = 4 A
Example 2:
Consider the circuit shown in Figure 4. Find the value of the current source current, Ia.
Figure 4 The circuit considered in Example 2.
Solution: The inputs to this circuit are the voltage of the independent voltage source and the
current of the independent current source. The response of this circuit is the voltage measured by
the voltmeter. The two inputs work together to produce a response that has a value of 9 V. This
circuit can be analyzed using superposition to separate the part of the response caused by the
voltage source from the part of the response caused by the current source.
Figure 5 shows the circuit from Figure 4 after replacing the voltmeter by an equivalent
open circuit and labeling the voltage measured by the voltmeter.
Figure 6a shows the circuit used to determine, v1, the part of the response caused by the
voltage source. The current source current is set to zero to determine the part of the response
caused by the voltage source. Consequently, the current source is replaced by an open circuit, i.e.
a zero current source, in Figure 6a.
Figure 6b shows the circuit used to determine, v2, the part of the response caused by the
current source. The voltage source voltage is set to zero to determine the part of the response
caused by the current source. Consequently, the voltage source is replaced by a short circuit, i.e.
a zero voltage source, in Figure 6b.
3
Figure 5 The circuit from Figure 4 after replacing the voltmeter by an open circuit.
(a)
(b)
Figure 6 The circuit from Figure 5 after using superposition. (a) The response to the voltage
source acting alone. (b) The response to the current source acting alone.
Apply voltage division to the circuit in Figure 3a to get
v1 =
2
6 =1 V
8+ 2+ 2
Use current division in Figure 3b to get
ib =
8
2
Ia = Ia
8 + ( 2 + 2)
3
Ohm’s Law gives
2  4
v2 = 2 ib = 2  I a  = I a
3  3
The response to both sources working together is equal to the sum of the responses to the
individual sources working separately. Consequently,
9 = v1 + v2 = 1 +
4
Ia
3
⇒ Ia = 6 A
4
Example 3:
Consider the circuit shown in Figure 7. Find the value of the current source current, Ia.
Figure 7 The circuit considered in Example 3.
Solution: The inputs to this circuit are the voltage of the independent voltage source and the
current of the independent current source. The response of this circuit is the current measured by
the ammeter. The two inputs work together to produce a response that has a value of 0.75 A.
This circuit can be analyzed using superposition to separate the part of the response caused by
the voltage source from the part of the response caused by the current source.
Figure 8 shows the circuit from Figure 7 after replacing the ammeter by an equivalent
short circuit and labeling the current measured by the ammeter.
Figure 9a shows the circuit used to determine, i1, the part of the response caused by the
voltage source. The current source current is set to zero to determine the part of the response
caused by the voltage source. Consequently, the current source is replaced by an open circuit, i.e.
a zero current source, in Figure 9a.
Figure 9b shows the circuit used to determine, i2, the part of the response caused by the
current source. The voltage source voltage is set to zero to determine the part of the response
caused by the current source. Consequently, the voltage source is replaced by a short circuit, i.e.
a zero voltage source, in Figure 9b.
Figure 8 The circuit from Figure 7 after replacing the ammeter by a short circuit.
5
(b)
(a)
Figure 9 The circuit from Figure 8 after using superposition. (a) The response to the voltage
source acting alone. (b) The response to the current source acting alone.
.
Apply KVL to the circuit in Figure 9a to get
2 − 6 i1 − 2 i1 = 0 ⇒ i1 =
1
A
4
Use current division in Figure 3b to get
i2 =
2
1
Ia = Ia
2+6
4
The response to both sources working together is equal to the sum of the responses to the
individual sources working separately. Consequently,
3
1 1
= i1 + i2 = + I a
4
4 4
⇒ Ia = 2 A
6
Example 4:
Consider the circuit shown in Figure 10. Find the value of the current source current, Ia.
Figure 10 The circuit considered in Example 4.
Solution: The inputs to this circuit are the voltage of the independent voltage source and the
current of the independent current source. The response of this circuit is the voltage measured by
the voltmeter. The two inputs work together to produce a response that has a value of 9 V. This
circuit can be analyzed using superposition to separate the part of the response caused by the
voltage source from the part of the response caused by the current source.
Figure 11 shows the circuit from Figure 10 after replacing the voltmeter by an equivalent
open circuit and labeling the voltage measured by the voltmeter.
Figure 12a shows the circuit used to determine, v1, the part of the response caused by the
voltage source. The current source current is set to zero to determine the part of the response
caused by the voltage source. Consequently, the current source is replaced by an open circuit, i.e.
a zero current source, in Figure 12a.
Figure 12b shows the circuit used to determine, v2, the part of the response caused by the
current source. The voltage source voltage is set to zero to determine the part of the response
caused by the current source. Consequently, the voltage source is replaced by a short circuit, i.e.
a zero voltage source, in Figure 12b.
Figure 11 The circuit from Figure 10 after replacing the voltmeter by a open circuit.
7
(b)
(a)
Figure 12 The circuit from Figure 11 after using superposition. (a) The response to the voltage
source acting alone. (b) The response to the current source acting alone.
Apply voltage division to the circuit in Figure 12a to get
v1 =
4
( −6 ) = −0.6 V
16 + 20 + 4
Use current division in Figure 3b to get
ib =
16
2
Ia = Ia
16 + ( 20 + 4 )
5
The Ohm’s Law gives
2  8
v2 = 4 ib = 4  I a  = I a = 1.6 I a
5  5
The response to both sources working together is equal to the sum of the responses to the
individual sources working separately. Consequently,
9 = v1 + v2 = −0.6 + 1.6 I a
⇒ Ia = 6 A
8