- The document is about using superposition to solve for currents and voltages in basic electric circuits. - In the first example circuit, superposition is used to find the total current ix as the sum of the currents with each source activated alone. This gives ix = 0.66 A. - In the second example, superposition is used to find the voltages across each current source. For source v1a it is found that v1a = 1.967 V, and for source v2a, v2a = -0.246 V. The total voltages are found to be v1 = 11.147 V and v2 = -1.394 V.

1. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com

2. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 2
S10_HWProblem1: Superposition
Note: This is Practice 5.1 on page 127, Figure 5.4 in the Ref 1 book (by Hayt)
 Use superposition to find ix : 0.66 A
2 = v1a /10 + (v1a - v2a )/15
60 = 3v1a + 2(v1a )
60 = 5 v1a
3.5 = v2b -ixb = 3.5 / 25 = 0.14 A
v2a = 0
v1a = 12 V ixa = 12/15 =0.8A
Short
circuit
(a)
v1a v2a
ixa
Open
circuit
(b)
v2bv1b
ixb
ixb = - 0.14A ix = ixa + ixb = 0.66 A
v1 v2
v2b / 25 = -ixb

3. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 3
S10_HWProblem2: Superposition
Note: This is Practice 5.2 on page 129, Figure 5.7 in the Ref 1 book (by Hayt)
 Use superposition to find voltage across each current source:
0 = (v1a - 3) / 7 + (v1a - v2a )/15
(a)
Open circuit
0 = 15v1a - 45 + 7v1a - 7v2a
45 = 22v1a- 7v2a
v2av1a
ia
4ia
4i a= v2a / 5 + (v2a - v1a )/15
3v2a / 5 = (v2a - v1a )/15
8v2a = - v1a v2a = - 0.246 V
v1a = 1.96 7 V
ia= v2a/5
KCL for v1a
KCL for v2a

4. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 4
S10_HWProblem2: Superposition
Note: This is Practice 5.2 on page 129, Figure 5.7 in the Ref 1 book (by Hayt)
 Use superposition to find voltage across each current source: v1 = 11.147 V
(a)
(b)(a)
v2av1a
Open circuit
Short circuit
4ia
ia
v2bv1b
4ib
ib
(b)
2 = v1b / 7 + (v1b - v2b )/15
4i b= v2b / 5 + (v2b - v1b )/15
210 = 22v1b - 7v2b
ib= v2b /5
8v2b = - v1b
v2b = - 1.148 V
v1b = 9.18 V
v1 = v1a + v1b v2 = v2a + v2b
v2 = -1.394 V
v2a = - 0.246 Vv1a = 1.96 7 V
KCL for v1b
KCL for v2b

5. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 5
References
Ref 1 Ref 2