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solutions_chapter19

I 5 1 60 3 10 15 2 e 100 s 5 1 60 3 10 15 21 160 3 10

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I 5 1 6.0 3 10 15 2 e 1.00 s 5 1 6.0 3 10 15 21 1.60 3 10 2 19 C 2 1.00 s 5 9.6 3 10 2 4 A 5 0.96 mA. 6.0 3 10 15 I 5 D Q D t . E I Current, Resistance, and Direct-Current Circuits 19-19
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19.78. Set Up: The power dissipated in a resistor is The power for an emf is When the current passes through an emf from the to the terminal, chemical energy is converted to electrical energy. When the current passes through the emf from the to the terminal, electrical energy is converted to chemical energy. Solve: (a) Assume the current in the circuit is counterclockwise. The loop rule gives: (b) (c) In chemical energy is being converted to electrical energy, at the rate 19.79. Set Up: Electrical energy is deposited in his body at the rate Heat energy Q produces a temperature change according to where Solve: (a) The energy deposited is Find when (b) An increase of only 63 brings the water in the body to the boiling point; part of the person’s body will be vaporized. Reflect: Even this approximate calculation shows that being hit by lightning is very dangerous. 19.80. Set Up: Solve: (a) The tube of seawater has resistance (b) The object and the remaining tube of seawater are in series. The 10 cm tube of seawater has resistance 130 . The object is 10 cm long, has the same cross-sectional area and half the resistance of seawater, so its resistance is 65 . The total equivalent resistance of the tube is The current now is The increase in current due to the proximity of the object is 19.81. Set Up: The network is sketched in Figure 19.81. If the current in is I, it is for and Since the current is greatest in that resistor dissipates the most power. Figure 19.81 Solve: so The total power dissipated therefore is Reflect: is dissipating its maximum of 32.0 W but and are dissipating only 8.0 W each. R 3 R 2 R 1 1 4.00 A 2 2 1 2.00 V 2 1 1 2.00 A 2 2 1 2.00 V 2 1 1 2.00 A 2 2 1 2.00 V 2 5 48.0 W. I 5 Å P R 5 Å 32.0 W 2.00 V 5 4.00 A. P 1 5 32.0 W I I / 2 I / 2 R 2 R 3 R 1 R 1 , P 5 I 2 R . R 3 . R 2 I / 2 R 1 15.4 mA 2 11.5 mA 5 3.9 mA. I 5 V R 5 3.0 V 195 V 5 15.4 mA. 130 V 1 65 V 5 195 V . V V I 5 V R 5 3.0 V 260 V 5 11.5 mA R 5 r L A 5 1 0.13 V # m 21 0.20 m 2 1.0 3 10 2 4 m 2 5 260 V . R 5 r L A . V 5 IR . D T 5 Q mc 5 2.5 3 10 7 J 1 75 kg 21 4190 J / kg # 2 5 80 C° Q 5 2.5 3 10 7 J. D T Pt 5 1 6.15 3 10 11 W 21 40 3 10 2 6 s 2 5 2.5 3 10 7 J. P 5 I 2 R 5 1 25,000 A 2 2 1 1.0 k V 2 5 6.25 3 10 11 W. c 5 4190 J / kg # C°. Q 5 mc D T , D T P 5 I 2 R . 1 12.0 V 21 0.40 A 2 5 4.8 W. E 1 P 5 I 2 R 1 I 2 r 1 1 I 2 r 2 5 I 2 1 R 1 r 1 1 r 2 2 5 1 0.40 A 2 2 1 8.0 V 1 1.0 V 1 1.0 V 2 5 1.6 W I 5 0.40 A. 12.0 V 2 8.0 V 2 I 1 1.0 V 1 8.0 V 1 1.0 V 2 5 0. 2 1 1 2 E I . I 2 R . 19-20 Chapter 19
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19.82. Set Up: Use a network of resistors that has Each resistor can dissipate less than 2.00 W but the total power is the sum of the powers for each resistor so it can be 2.00 W. Solve: (a) and (b) One possible network is two in parallel in series with two in parallel, as shown in Figure 19.82. Each parallel combination has an equivalent resistance of so these two units in series have an equivalent resist- ance of The current is the same in each resistor so each resistor dissipates the same power. If each resistor dissipates 0.50 W, then the network dissipates 2.00 W. Other combinations also give the desired resistance and power
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