19_InstSolManual_PDF_Part23

19_InstSolManual_PDF_Part23 - and V a 2 V b 5 2 2.0 V 1 1...

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Reflect: The parallel combination has less equivalent resistance even though both cables contain the same volume of each metal. 19.88. Set Up: Because of the polarity of each emf, the current in the resistor must be in the direction shown in Figure 19.88. Let I be the current in the 24.0 V battery. Figure 19.88 Solve: The loop rule applied to loop (1) gives: The junction rule then says that the current in the middle branch is 2.00 A, as shown in Figure 19.88b. The loop rule applied to loop (2) gives: and 19.89. Set Up: There is no current in the middle branch because there is not a complete conducting path for that branch. There is only a single current in the circuit, as shown in Figure 19.89. To find start at point b and travel to point a. Figure 19.89 Solve: Going from a to b through the 12.0 V battery gives Or, going from a to b through the 8.0 V battery gives
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Unformatted text preview: and V a 2 V b 5 2 2.0 V 1 1 0.444 A 21 5.00 V 2 5 1 0.22 V. V a 2 I 1 2.00 V 2 2 I 1 1.00 V 2 2 8.0 V 2 I 1 2.00 V 2 1 10.0 V 5 V b V a 2 V b 5 12.0 V 2 10.0 V 2 1 0.444 A 21 4.00 V 2 5 1 0.22 V. V a 1 I 1 2.00 V 2 1 I 1 1.00 V 2 2 12.0 V 1 I 1 1.00 V 2 1 10.0 V 5 V b . I 5 0.444 A. 12.0 V 2 8.0 V 2 I 1 2.00 V 1 2.00 V 1 1.00 V 1 2.00 V 1 1.00 V 1 1.00 V 2 5 0. I a b I I I 12.0 V 10.0 V 1.00 V 8.0 V 1.00 V 1.00 V 3.00 V 2.00 V 2.00 V 2.00 V 1.00 V + + + V ab 5 V a 2 V b , E 5 8.6 V. 1 E 2 1 1.80 A 21 7.00 V 2 1 1 2.00 A 21 2.00 V 2 5 I 5 3.80 A. 1 24.0 V 2 1 1.80 A 21 7.00 V 2 2 I 1 3.00 V 2 5 0. I 24.0 V 24.0 V 3.80 A 2.00 A 1.8 A 1.80 A 7.00 V 2.00 V 3.00 V 7.00 V 2.00 V 3.00 V 1 1 2 1 2 2 ( a ) ( b ) + + + + E E 7.00 V Current, Resistance, and Direct-Current Circuits 19-23...
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