To determine Calculate the power developed in the dependent voltage source in the given circuit using PSPICE. Explanation The given circuit is redrawn as shown in Figure 1. Apply Kirchhoff’s voltage law for Loop 1 in Figure 1. 3 ( i 1 − i 2 ) + 5 ( i 1 − i 3 ) = − 53 i Δ 3 i 1 − 3 i 2 + 5 i 1 − 5 i 3 = − 53 i Δ 8 i 1 − 3 i 2 − 5 i 3 = − 53 i Δ (1) Apply Kirchhoff’s voltage law for Loop 2 in Figure 1. 3 ( i 2 − i 1 ) + 20 ( i 2 − i 3 ) + 7 i 2 = 30 3 i 2 − 3 i 1 + 20 i 2 − 20 i 3 + 7 i 2 = 30 − 3 i 1 + 30 i 2 − 20 i 3 = 30 (2) Apply Kirchhoff’s voltage law for Loop 3 in Figure 1. 5 ( i 3 − i 1 ) + 20 ( i 3 − i 2 ) + 2 i 3 = 30 5 i 3 − 5 i 1 + 20 i 3 − 20 i 2 + 2 i 3 = 30 − 5 i 1 − 20 i 2 + 27 i 3 = 30 (3) In Figure 1, the current i Δ is calculated as follows. i Δ = i 2 − i 3 (4) Substitute equation (4) in equation (1). 8 i 1 − 3 i 2 − 5 i 3 = − 53 ( i 2 − i 3 ) 8 i 1 − 3 i 2 − 5 i 3 = − 53 i 2 + 53 i 3 8 i 1 − 3 i 2 − 5 i 3 + 53 i 2 − 53 i 3 = 0 8 i 1 + 50 i 2 − 58 i 3 = 0 Rearrange the equation as follows, 58 i 3 = 8 i 1 + 50 i 2 i 3 = 8 i 1 + 50 i 2 58 i 3 = 0.1379 i 1 + 0.8621 i 2 (5) Substitute equation (5) in equation (2). − 3 i 1 + 30 i 2 − 20 ( 0.1379 i 1 + 0.8621 i 2 ) = 30 − 3 i 1 + 30 i 2 − 2.758 i 1 − 17.242 i 2 = 30 − 5.758 i 1 + 12.758 i 2 = 30 Rearrange the equation as follows, 5.758 i 1 = 12.758 i 2 − 30 i 1 = 12.758 i 2 − 30 5.758 i 1 = 2.2157 i 2 − 5.2101 (6) Substitute equation (6) in equation (5)

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ISBN: 9780134747224

Author: Riedel

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 4, Problem 40P

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Chapter 4, Problem 39P

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