19_InstSolManual_PDF_Part14

19_InstSolManual_PDF_Part14 - 5 42.0 V. 2 1 4.00 A 21 6.00...

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Loop (3): reads reads reads reads Reflect: The sum of potential changes around loop (4) is The loop rule is satisfied for loop (4) and this is a good check of our calculations. 19.60. Set Up: The circuit diagram is given in Figure 19.60. The junction rule has been used to find the magnitude and direction of the current in the upper branch of the circuit. There are no remaining unknown currents. Figure 19.60 Solve: (a) The junction rule gives that the current in R is 2.00 A to the left. (b) The loop rule applied to loop (1) gives: (c) The loop rule applied to loop (2) gives: E 5 24.0 V 1 18.0 V
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Unformatted text preview: 5 42.0 V. 2 1 4.00 A 21 6.00 V 2 1 E 2 1 6.00 A 21 3.00 V 2 5 0. R 5 28.0 V 2 18.0 V 2.00 A 5 5.00 V . 2 1 2.00 A 2 R 1 28.0 V 2 1 6.00 A 21 3.00 V 2 5 0. R 6.00 V 3.00 V 28.0 V E 1 2 2 2.00 A 4.00 A 6.00 A + + 1 1 2 200.0 V 2 200.0 V 1 160.0 V 2 160.0 V 5 0. 1 8.00 A 21 20.0 V 2 5 200.0 V 2 1 10.0 A 21 20.0 V 2 1 1 4.00 A 21 40.0 V 2 2 1 200.0 V 2 I 1 R 1 I 2 1 40.0 V 2 2 I 5 1 20.0 V 2 5 I 5 5 8.00 A. A 5 I 1 1 I 2 5 14.0 A. A 4 I 2 1 I 5 5 12.0 A. A 3 I 2 5 4.00 A. A 2 I 5 5 160.0 V 20.0 V 5 8.00 A 1 160.0 V 2 I 5 1 20.0 V 2 5 0. 19-14 Chapter 19...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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