Basic Electric Circuits Session 10 Homework solutions
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- 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.
Basic Electric Circuits Session 10 Homework solutions
- 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