19_InstSolManual_PDF_Part15

19_InstSolManual_PDF_Part15 - ). Figure 19.62 10.0 V 1.0 V...

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19.61. Set Up: We can use the power consumption in the 5.00 resistor to find the current through it. The circuit, unknown currents, and a circuit loop are all shown in Figure 19.61. Figure 19.61 Solve: so The loop rule for loop (1) gives The ammeter reads 0.714 A. 19.62. Set Up: so the power consumption of the resistor allows us to calculate the current through it. Unknown currents and are shown in Figure 19.62. The junction rule says that In Figure 19.62 the two resistors in parallel have been replaced by their equivalent (
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Unformatted text preview: ). Figure 19.62 10.0 V 1.0 V I 3 17 V 13 V 19 V 6.0 V 3.0 V 25 V E I 1 I 2 1 1 2 1 2 2 + + 10.0 V 20.0 V I 1 5 I 2 1 I 3 . I 3 I 2 I 1 , 6.0 V P 5 I 2 R I 2 5 15.0 V 2 10.0 V 7.00 V 5 0.714 A. 1 15.0 V 2 I 2 1 7.00 V 2 2 1 2.00 A 21 5.00 V 2 5 0. I 1 5 Å 20.0 W 5.00 V 5 2.00 A. P 5 I 2 R 15.0 V E 7.00 V 2.00 V 5.00 V I 1 1 1 2 + + I 2 V Current, Resistance, and Direct-Current Circuits 19-15...
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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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