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Transcript
Homework 4 Ch18: Q 7; P 7, 9, 19, 39; Ch19: Q1; P17, 37, 45 Ch18 Question: 7. If the resistance of a small immersion heater (to heat water for tea or soup, Fig. 18– 32) was increased, would it speed up or slow down the heating process? Explain. Solution We assume that the voltage is the same in both cases. Then if the resistance increases, the power delivered to the heater will decrease according to P V 2 R . If the power decreases, the heating process will slow down. Problems: 7. (II) An electric clothes dryer has a heating element with a resistance of 9.6 . (a) What is the current in the element when it is connected to 240 V? (b) How much charge passes through the element in 50 min? Solution (a) Use Eq. 18-2b to find the current. V IR I (b) R 240 V 9.6 25 A Use the definition of current, Eq. 18-1. I 9. V Q t Q I t 25 A 50 min 60 s min 7.5 10 4 C (II) A bird stands on a dc electric transmission line carrying 2800 A (Fig. 18–34). The line has 2.5 105 resistance per meter, and the bird’s feet are 4.0 cm apart. What is the potential difference between the bird’s feet? Solution Find the potential difference from the resistance and the current. R 2.5 105 m 4.0 102 m 1.0 10 6 V IR 2800 A 1.0 106 2.8 103 V *19. (II) A 100-W lightbulb has a resistance of about 12 when cold (20°C) and 140 when on (hot). Estimate the temperature of the filament when hot assuming an average temperature coefficient of resistivity 0.0060 (Cº ) 1. Solution R R0 1 T T0 T T0 1 R 1 140 o o o 1 20 C 12 1 1798 C 1800 C 1 o R0 0.0060 C 39. (II) A power station delivers 620 kW of power at 12,000 V to a factory through wires with total resistance 3.0 . How much less power is wasted if the electricity is delivered at 50,000 V rather than 12,000 V? Solution Find the current used to deliver the power in each case, and then find the power dissipated in the resistance at the given current. P IV I P V Pdissipated = I R 2 ; 12,000 V 50,000 V R 2 2 4 2 5 Pdissipated V2 6.20 10 W 3.0 8008 W 1.2 10 V 6.20 10 W 3.0 461W 5 10 V 5 Pdissipated P2 4 2 difference 8008 W 461W 7.5 103 W Ch19 Question: 1. Explain why birds can sit on power lines safely, whereas leaning a metal ladder up against a power line to fetch a stuck kite is extremely dangerous. Solution The birds are safe because they are not grounded. Both of their legs are essentially at the same voltage (the only difference being due to the small resistance of the wire between their feet), and so there is no current flow through their bodies since the potential difference across their legs is very small. If you lean a metal ladder against the power line, you are making essentially a short circuit from the high potential wire to the low potential ground. A large current will flow at least momentarily, and that large current will be very dangerous to anybody touching the ladder. Problems: 17. (II) Determine (a) the equivalent resistance of the circuit shown in Fig. 19–39, and (b) the voltage across each resistor. Solution (a) The equivalent resistance is found by combining the 820 and 680 resistors in parallel, and then adding the 470 resistor in series with that parallel combination. 1 1 1 Req 470 372 470 842 840 820 680 (b) The current delivered by the battery is I V Req 12.0 V 842 1.425 102 A . This is the current in the 470 resistor. The voltage across that resistor can be found by Ohm’s law V470 IR 1.425 102 A 470 6.7 V . Thus the voltage across the parallel combination must be 12.0 V 6.7 V 5.3 V . This is the voltage across both the 820 and 680 resistors, since parallel resistors have the same voltage across them. Note that this voltage value could also be found as follows. Vparallel IRparallel 1.425 10 2 A 372 5.3 V 37. (I) A 3.00-F and a 4.00-F capacitor are connected in series, and this combination is connected in parallel with a 2.00-F capacitor (see Fig. 19–52). What is the net capacitance? Solution The series capacitors add reciprocally, and then the parallel combination is found by adding linearly. 1 1 1 Ceq C3 C1 C2 45. 1 1 1 6 6 2.00 10 F 3.71 10 F 3.71 F 6 6 3.00 10 F 4.00 10 F (II) A 0.40-F and a 0.60-F capacitor are connected in series to a 9.0-V battery. Calculate (a) the potential difference across each capacitor, and (b) the charge on each. (c) Repeat parts (a) and (b) assuming the two capacitors are in parallel. Solution When the capacitors are connected in series, they each have the same charge as the net capacitance. 1 (a) 1 1 1 1 Q1 Q2 Qeq CeqV V 6 6 0.40 10 F 0.60 10 F C1 C2 1 9.0 V 2.16 106 C V1 Q1 C1 2.16 106 C 0.40 106 F 5.4 V V2 Q2 C2 2.16 10 6 C 0.60 10 6 F 3.6 V (b) Q1 Q2 Qeq 2.16 106 C 2.2 106 C When the capacitors are connected in parallel, they each have the full potential difference. (c) V1 9.0 V V2 9.0 V Q1 C1V1 0.40 106 F 9.0 V 3.6 106 C Q2 C2V2 0.60 106 F 9.0 V 5.4 106 C