To determine Find the total power dissipated in the given circuit using PSPICE. Explanation The given circuit is redrawn with reference node and the node voltages v 1 , v 2 and v 3 as shown in Figure 1. Apply Kirchhoff’s current law at node v 1 in Figure 1. v 1 + 40 12 + v 1 − v 2 20 + v 1 25 = − 5 v 1 12 + 40 12 + v 1 20 − v 2 20 + v 1 25 = − 5 v 1 ( 1 12 + 1 20 + 1 25 ) − v 2 20 = − 5 − 40 12 0.1733 v 1 − 0.05 v 2 = − 8.3333 (1) Apply Kirchhoff’s current law at node v 2 in Figure 1. v 2 − v 1 20 + v 2 − v 3 40 = 5 + 7.5 v 2 20 − v 1 20 + v 2 40 − v 3 40 = 12.5 − v 1 20 + v 2 ( 1 20 + 1 40 ) − v 3 40 = 12.5 − 0.05 v 1 + 0.075 v 2 − 0.025 v 3 = 12.5 (2) Apply Kirchhoff’s current law at node v 3 in Figure 1. v 3 40 + v 3 − v 2 40 = − 7.5 v 3 40 + v 3 40 − v 2 40 = − 7.5 − v 2 40 + v 3 ( 1 40 + 1 40 ) = − 7.5 − 0.025 v 2 + 0.05 v 3 = − 7.5 (3) Rearrange the equation (1) as follows, 0.1733 v 1 = − 8.3333 + 0.05 v 2 v 1 = − 8.3333 + 0.05 v 2 0.1733 v 1 = − 48.0859 + 0.2885 v 2 (4) Substitute the equation (4) in equation (2). − 0.05 ( − 48.0859 + 0.2885 v 2 ) + 0.075 v 2 − 0.025 v 3 = 12.5 2.4043 − 0.0144 v 2 + 0.075 v 2 − 0.025 v 3 = 12.5 0.0606 v 2 − 0.025 v 3 = 12.5 − 2.4043 0.0606 v 2 − 0.025 v 3 = 10.0957 Rearrange the equation as follows, 0.0606 v 2 = 10.0957 + 0.025 v 3 v 2 = 10.0957 + 0.025 v 3 0.0606 v 2 = 166.5957 + 0.4125 v 3 (5) Substitute the equation (5) in equation (3). − 0.025 ( 166.5957 + 0.4125 v 3 ) + 0.05 v 3 = − 7.5 − 4.1648 − 0.0103 v 3 + 0.05 v 3 = − 7.5 0.0397 v 3 = − 7.5 + 4.1648 0.0397 v 3 = − 3.3352 Rearrange the equation as follows, v 3 = − 3.3352 0.0397 = − 84.01 V ≅ − 84 V Substitute − 84 for v 3 in equation (3) to find the voltage v 2 in volts. − 0.025 v 2 + 0.05 ( − 84 ) = − 7.5 − 0.025 v 2 − 4.2 = − 7.5 − 0.025 v 2 = − 7.5 + 4.2 − 0.025 v 2 = − 3.3 Rearrange the equation as follows, v 2 = − 3.3 − 0.025 = 132 V Substitute 132 for v 2 in equation (4) to find the voltage v 1 in volts. v 1 = − 48.0859 + 0.2885 ( 132 ) = − 10 V In Figure 1, the power dissipated in the 12 ohms resistor is calculated as follows. p 12 Ω = ( v 1 + 40 ) 2 12 Substitute − 10 for v 1 to find the power p 12 Ω in watts. p 12 Ω = ( − 10 + 40 ) 2 12 = 30 2 12 = 75 W The power dissipated in the 25 ohms resistor is calculated as follows. p 25 Ω = v 1 2 25 Substitute − 10 for v 1 to find the power p 25 Ω in watts

11th Edition

ISBN: 9780134746968

Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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Chapter 4, Problem 15P

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Find the total power dissipated in the given circuit using PSPICE.

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