solutions_chapter19

(c the sum of the potential changes around the

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Unformatted text preview: (c) The sum of the potential changes around the circuit must be zero, so if there is a potential rise of 12.0 V across the battery, there must be a potential drop of 12.0 V across the switch. The voltmeter reads 12.0 V. (d) and (a) The terminal voltage of the battery is (b) The voltage across the resistor is (c) The meter is now a path of zero resistance and the voltage across it is zero. 19.29. Set Up: The terminal voltage of the battery is given by The terminal voltage also equals the voltage across the resistor, so We have two unknowns so we need two equations. Solve: (a) With the 1500 M resistor, gives With the resistor, gives Subtracting the second equation from the first gives and (b) Now R is 7.00 . Or, Reflect: The smaller R is, the larger the current and the smaller the terminal voltage. With the 1500 M resistor the current is very small and the terminal voltage differs only very little from the emf of the battery. 19.30. Set Up: When a battery is short-circuited, Solve: (a) (b) (c) r 5 6.00 V 1000 A 5 6.00 3 10 2 3 V 5 6.00 m V r 5 1.50 V 10.0 A 5 0.150 V r 5 E I 5 1.50 V 25.0 A 5 0.0600 V E 2 Ir 5 0. V V ab 5 IR 5 1.91 V. V ab 5 E 2 Ir 5 2.50 V 2 1 0.274 A 21 2.14 V 2 5 1.91 V. I 5 E R 1 r 5 2.50 V 7.00 V 1 2.14 V 5 0.274 A V E 5 1.75 V 1 1 0.350 A 2 r 5 2.50 V r 5 2.14 V . 0.75 V 5 1 0.350 A 2 r 1.75 V 5 E 2 1 0.350 A 2 r . V ab 5 E 2 Ir I 5 V ab R 5 1.75 V 5.00 V 5 0.350 A. 5.00 V 2.50 V 5 E 2 1 1.67 3 10 2 9 A 2 r . V ab 5 E 2 Ir I 5 V ab R 5 2.50 V 1500 3 10 6 V 5 1.67 3 10 2 9 A. V 1 E , r 2 V ab 5 IR . V ab 5 E 2 Ir . 11.3 V. IR 5 1 2.40 A 21 4.70 V 2 5 E 2 Ir 5 12.0 V 2 1 2.40 A 21 0.30 V 2 5 11.3 V. I 5 E r 1 R 5 12.0 V 4.70 V 1 0.30 V 5 2.40 A. E 2 Ir 2 IR 5 V 5 IR I 5 0, I 5 0, R 5 V I 5 2.97 V 1.65 A 5 1.80 V . V 2 IR 5 r 5 V 2 E I 5 3.08 V 2 2.97 V 1.65 A 5 0.067 V . I 5 1.65 A, V 5 2.97 V V 5 E 2 Ir , E 5 3.08 V. V 5 E 2 Ir . E . 15.0 A . I T 5 15.0 A 1 1 1 0.0020 1 C° 2 2 1 21 100 C° 2 5 12.5 A. I 5 T 5 100°C T 5 0°C, I T 5 I 1 1 a 1 T 2 T 2 . I 5 V R R T 5 R 3 1 1 a 1 T 2 T 24 . I new 5 V p d new 2 4 r L new 5 V p 1 2 d 2 2 4 r 1 2 L 2 5 2 I d new 5 2 d . L new 5 2 L ; Current, Resistance, and Direct-Current Circuits 19-5 19.31. Set Up: When current passes through a battery in the direction from the terminal toward the terminal, the terminal voltage of the battery is Solve: (a) gives (b) so Reflect: The total resistance in the circuit is which agrees with the value specified in the problem. 19.32. Set Up: The graph shows when Solve: (a) When (b) When As A qualitative sketch of versus R is sketched in Figure 19.32a. Figure 19.32 (c) If then The graph of I versus R is sketched in Figure 19.32b. when and when is constant, independent of R. The graph of versus R is sketched in Figure 19.32c....
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(c The sum of the potential changes around the circuit must...

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