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Solutions to Problems 1. Use the definition of current, Eq. 18-1. Q 1.30 C 1 electron I 1.30 A 8.13 1018 electrons s 19 t s 1.60 10 C 2. Use the definition of current, Eq. 18-1. Q I Q I t 6.7 A 5.0 h 3600 s h 1.2 105 C t 3. Use the definition of current, Eq. 18-1. 19 Q 1200 ions 1.60 10 C ion I 5.5 1011 A 6 t 3.5 10 s 4. Use Eq. 18-2a for resistance. V 120 V R 29 I 4.2 A 5. Use Eq. 18-2b for the voltage. V IR 0.25 A 3800 950 V 6. (a) (b) 7. (a) (b) 8. Use Eq. 18-2a for resistance. V 120 V R 16 I 7.5 A Use the definition of current, Eq. 18-1. Q I Q I t 7.5 A 15 min 60 s min 6.8 103 C t Use Eq. 18-2b to find the current. V 240 V V IR I 25 A R 9.6 Use the definition of current, Eq. 18-1. Q I Q I t 25 A 50 min 60 s min 7.5 10 4 C t Find the current from the voltage and resistance, and then find the number of electrons from the current. V 9.0 V V IR I 5.625 A R 1.6 5.625 A 5.625 C 1 electron 19 60s s 1.60 10 C 1min 2.1 1021 electrons min © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 47 Physics: Principles with Applications, 6th Edition Giancoli 9. 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 10. (a) If the voltage drops by 15%, and the resistance stays the same, then by Eq. 18-2b, V IR , the current will also drop by 15%. I final 0.85 I initial 0.85 6.50 A 5.525 A 5.5 A (b) If the resistance drops by 15% (the same as being multiplied by 0.85), and the voltage stays the same, then by Eq. 18-2b, the current must be divided by 0.85. I 6.50 A I final initial 7.647 A 7.6 A 0.85 0.85 11. (a) (b) Use Eq. 18-2a to find the resistance. V 12 V R 20 I 0.60 A An amount of charge Q loses a potential energy of Q V as it passes through the resistor. The amount of charge is found from Eq. 18-1. PE Q V I t V 0.60 A 60 s 12 V 430 J 12. Use Eq. 18-3 to find the diameter, with the area as A r 2 d 2 4 . R 13. A 4L d 2 d 4L R 4 1.00 m 5.6 108 m 0.32 L A 4L d 2 1.68 108 m 4 3.5 m 1.5 10 m 4 m 3 2 3.3 102 Use Eq. 18-3 to calculate the resistances, with the area as A r 2 d 2 4 , and so L 4L R . A d2 4L Al Al2 2 2.65 108 m 10.0 m 2.5 mm RAl d Al Al LAl d Cu2 1.2 2 8 4 LCu Cu LCu d Al2 RCu 1.68 10 m 20.0 m 2.0 mm Cu 2 d Cu 15. 4.7 10 Use Eq. 18-3 to calculate the resistance, with the area as A r 2 d 2 4 . R 14. L Use Eq. 18-3 to express the resistances, with the area as A r 2 d 2 4 , and so © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 48 Physics: Principles with Applications, 6th Edition Giancoli R L A 4L d2 . RW RCu W 4L d 2 W Cu 4L d Cu2 W 5.6 108 m 2.5 mm 4.6 mm Cu 1.68 108 m The diameter of the tungsten should be 4.6 mm. d W d Cu 16. Since the resistance is directly proportional to the length, the length of the long piece must be 4.0 times the length of the short piece. L Lshort Llong Lshort 4.0 Lshort 5.0 Lshort Lshort 0.20 L , Llong 0.80 L Make the cut at 20% of the length of the wire . Lshort 0.20 L , Llong 0.80 L Rshort 0.2 R 2.0 , Rlong 0.8R 8.0 17. Use Eq. 18-4 multiplied by L A so that it expresses resistance instead of resistivity. R R0 1 T T0 1.15R0 1 T T0 1.15 T T0 0.15 0.15 .0068 C o 1 22 Co So raise the temperature by 22 Co to a final temperature of 42 o C . 18. Use Eq. 18-4 for the resistivity. T Cu 0 Cu 1 Cu T T0 0 W T T0 19. 1 0 W Cu 0 Cu 1 20o C 5.6 108 m 1 363.1o C 360o C o 1 1.68 10 8 m 0.0068 C 1 Use Eq. 18-4 multiplied by L A so that it expresses resistances instead of resistivity. R R0 1 T T0 T T0 1 R 1 140 o 1 1798o C 1800o C 1 20 C 1 o R0 0.0060 C 12 20. Calculate the voltage drop by combining Ohm’s Law (Eq. 18-2b) with the expression for resistance, Eq. 18-3. 4 1.68 108 m 26 m L 4 L V IR I I 12 A 2.5 V 2 A d2 1.628 103 m 21. In each case calculate the resistance by using Eq. 18-3 for resistance. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 49 Physics: Principles with Applications, 6th Edition Giancoli 22. 5 2 Lx 3.0 10 m 1.0 10 m 3.75 104 3.8 104 2 2 Ayz 2.0 10 m 4.0 10 m (a) Rx (b) Ry (c) Lz Rz Axy Ly Axz 3.0 10 m 2.0 10 m 1.5 10 1.0 10 m 4.0 10 m 3.0 10 m 4.0 10 m 6.0 10 1.0 10 m 2.0 10 m 5 2 2 3 3 2 5 2 2 2 The wires have the same resistance and the same resistivity. Rlong Rshort Llong A1 Lshort A2 4 2Lshort dlong 2 4 Lshort d 2 short dlong dshort 2 The original resistance is R0 V I 0 , and the high temperature resistance is R V I , where the 23. two voltages are the same. The two resistances are related by Eq. 18-4, multiplied by L A so that it expresses resistance instead of resistivity. 1 R 1V I 1I R R0 1 T T0 T T0 1 T0 1 T0 0 1 R0 V I0 I 20.0o C 1 0.00429 C o 1 0.4212 A o 0.3618 A 1 58.3 C 24. (a) Calculate each resistance separately using Eq. 18-3, and then add the resistances together to find the total resistance. RCu 8 Cu L 4 Cu L 4 1.68 10 m 5.0 m 0.10695 2 A d2 1.0 103 m RAl 8 Al L 4 Al L 4 2.65 10 m 5.0 m 0.16870 2 A d2 1.0 103 m Rtotal RCu RAl 0.10695 0.16870 0.27565 0.28 (b) (c) The current through the wire is the voltage divided by the total resistance. V 85 103 V I 0.30836 A 0.31A Rtotal 0.27565 For each segment of wire, Ohm’s Law is true. Both wires have the current found in (b) above. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 50 Physics: Principles with Applications, 6th Edition Giancoli VCu IRCu 0.30836 A 0.10695 0.033 V VAl IRAl 0.30836 A 0.16870 0.052 V Notice that the total voltage is 85 mV. 25. of The total resistance is to be 4700 ohms Rtotal at all temperatures. Write each resistance in terms Eq. 18-4 (with T0 0 C ), multiplied by L A to express resistance instead of resistivity. o Rtotal R0C 1 CT R0N 1 NT R0C R0C CT R0N R0N NT R0C R0N R0C C R0N N T For the above to be true, the terms with a temperature dependence must cancel, and the terms without a temperature dependence must add to Rtotal . Thus we have two equations in two unknowns. R0C C 0 R0C C R0 N N T R0 N R0C C Rtotal R0C R0 N R0C N N R0C N C N 0.0004 Co N R0C Rtotal 4700 2100 1 1 N C 0.0004 Co 0.0005 Co 1 R0 N Rtotal R0C 2600 26. Use Eq. 18-6b to find the resistance from the voltage and the power. P 27. V2 R R V2 P 240 V 2 3300 W 17 Use Eq. 18-5 to find the power from the voltage and the current. P IV 0.32 A 3.0 V 0.96 W 28. Use Eq. 18-6b to find the voltage from the power and the resistance. V2 P V RP 2700 0.25 W 26 V R 29. Use Eq. 18-6b to find the resistance, and Eq. 18-5 to find the current. (a) P V2 R R P IV I (b) P V2 R V2 P P V 120 V 2 75 W 75 W 120 V R 192 190 0.625 A 0.63 A V2 P 120 V 2 440 W 32.73 33 © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 51 Physics: Principles with Applications, 6th Edition Giancoli P IV I 30. (a) (b) (a) V 440 W 3.667 A 3.7 A 120 V Use Eq. 18-5 to find the current. P 110 W P IV I 0.96 A V 115 V Use Eq. 18-6b to find the resistance. P 31. P Since P V2 V2 R R R R V2 V2 P P 115 V 2 110 W 120 says that the resistance is inversely proportional to the power for a constant voltage, we predict that the 850 W setting has the higher resistance. 32. (b) R (c) R V2 P V2 P 120 V 2 850 W 120 V 2 1250 W 17 12 The power (and thus the brightness) of the bulb is proportional to the square of the voltage, V2 according to Eq. 18-6a, P . Since the resistance is assumed to be constant, if the voltage is cut R in half from 240 V to 120V, the power will be reduced by a factor of 4. Thus the bulb will appear only about 1/4 as bright in the United States as in Europe. 33. To find the kWh of energy, multiply the kilowatts of power consumption by the number of hours in operation. 1kW 1h 15 min 0.1375 kWh 0.14 kWh 1000 W 60 min Energy P in kW t in h 550 W To find the cost of the energy used in a month, multiply times 4 days per week of usage, times 4 weeks per month, times the cost per kWh. kWh 4 d 4 week 9.0 cents Cost 0.1375 20 cents month d 1week 1month kWh 34. To find the cost of the energy, multiply the kilowatts of power consumption by the number of hours in operation times the cost per kWh. 1kW 24 h $0.095 Cost 25 W 365day $ 21 1000 W 1day kWh 35. (a) Calculate the resistance from Eq. 18-2a, and the power from Eq. 18-5. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 52 Physics: Principles with Applications, 6th Edition Giancoli R V I 3.0 V 6.667 6.7 P IV 0.45 A 3.0 V 1.35 W 1.4 W 0.45 A If four D-cells are used, the voltage will be doubled to 6.0 V. Assuming that the (b) resistance of the bulb stays the same (by ignoring heating effects in the filament), the power that the bulb V2 would need to dissipate is given by Eq. 18-6b, P . A doubling of the voltage means the R power is increased by a factor of 4 . This should not be tried because the bulb is probably not rated for such a high wattage. The filament in the bulb would probably burn out, and the glass bulb might even explode if the filament burns violently. 36. The A h rating is the amount of charge that the battery can deliver. The potential energy of the charge is the charge times the voltage. 3600s PE QV 85 A h 12 V 3.7 106 J 1h 37. Each bulb will draw an amount of current found from Eq. 18-5. P P IV I bulb V The number of bulbs to draw 15 A is the total current divided by the current per bulb. VI P 120 V 15 A I total nI bulb n n total 18 bulbs V P 100 W 38. Find the power dissipated in the cord by Eq. 18-6a, using Eq. 18-3 for the resistance. 8 2.7 m 2L 2 P I 2R I 2 15.0 A 1.68 108 m 16 W 2 A 0.129 102 m 39. 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 P2 P IV I Pdissipated = I 2 R 2 R V V 6.20 10 W 3.0 8008 W 1.2 10 V 2 5 Pdissipated 12,000 V 4 6.20 10 W 3.0 461W 5 10 V 2 5 Pdissipated 50,000 V 2 4 difference 8008 W 461W 7.5 103 W 2 40. The water temperature rises by absorbing the heat energy that the electromagnet dissipates. Express both energies in terms of power, which is energy per unit time. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 53 Physics: Principles with Applications, 6th Edition Giancoli Pelectric Pto heat IV Qheat water t water m t IV cT mcT 17.5 A 240 V 4186 J kg C 7.50 C o t o 0.134 kg s 0.13 kg s This is about 134 ml per second. 41. (a) By conservation of energy and the efficiency claim, 60% of the electrical power dissipated by the heater must be the rate at which energy is absorbed by the water. Q mcT 0.6 Pemitted by Pabsorbed 0.6 IV heat water t t electromagnet by water I (b) 42. mcT 0.120 kg 4186 J kg 95o C 25o C 10.17 A 10 A 0.6 12 V 480s 0.6Vt Use Ohm’s Law to find the resistance of the heater. V 12 V V IR R 1.2 I 10.17 A Use Ohm’s law and the relationship between peak and rms values. V 220 V I peak 2 I rms 2 rms 2 0.14 A R 2.2 103 43. Find the peak current from Ohm’s law, and then find the rms current from the relationship between peak and rms values. I peak Vpeak R 180 V 330 0.54545 A I rms I peak 0.55 A 2 0.54545 A 2 0.39 A 44. (a) When everything electrical is turned off, no current will be flowing into the house, even though a voltage is being supplied. Since for a given voltage, the more resistance, the lower the current, a zero current corresponds to an infinite resistance. (b) Assuming that the voltage is 120 V, use Eq. 18-6a to calculate the resistance. P V2 R R V2 P = 120 V 2 75 W 192 1.9 102 45. The power and current can be used to find the peak voltage, and then the rms voltage can be found from the peak voltage. I peak 2 1500 W 2P P I rmsVrms Vrms Vrms 3.9 102 V I peak 5.4 A 2 46. Use the average power and rms voltage to calculate the peak voltage and peak current. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 54 Physics: Principles with Applications, 6th Edition Giancoli (a) Vpeak 2Vrms 2 660 V 933.4 V 9.3 102 V (b) P I rmsVrms I peak 2 Vrms I peak 2P Vrms 2 1800 W 660 V 3.9 A 47. (a) We assume that the 3.0 hp is the average power, so the maximum power is twice that, or 6.0 hp, as seen in Figure 18-22. 746 W 6.0 hp 4476 W 4.5 103 W 1hp (b) Use the average power and the rms voltage to find the peak current. P I rmsVrms 48. 9c. (a) I peak 2 Vrms I peak 2P Vrms 2 12 4476 W 240 V 13A The average power used can be found from the resistance and the rms voltage by Eq. 18- P 2 Vrms 240 V 2 1694 W 1.7 103 W R 34 (b) The maximum power is twice the average power, and the minimum power is 0. Pmax 2 P 2 1694 W 3.4 103 W 49. We follow exactly the derivation in Example 18-14, which results in an expression for the drift velocity. I I 4I m vd 2 neA N 1 mole N D e d 2 D e 12 d m 1 mole m 1.60 10 4 2.3 10 6 A 63.5 10 3 kg 6.02 10 8.9 10 23 50. Pmin 0 W 3 kg 3 19 C 0.65 10 m 3 2 5.1 1010 m s Use Ohm’s Law to find the resistance. V 22.0 103 V V IR R 0.02933 0.029 I 0.75 A Find the resistivity from Eq. 18-3. (a) (b) R L RA R r 2 0.02933 1.0 103 m 5.80 m A L L Find the number of electrons per unit volume from Eq. 18-10. (c) 2 1.6 108 m I neAvd n I eAvd I e r vd 2 1.60 10 19 0.75 A C 1.0 10 m 3 1.7 10 2 5 ms 8.8 1028 e m 3 © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 55 Physics: Principles with Applications, 6th Edition Giancoli 51. We are given a charge density and a speed (like the drift speed) for both types of ions. From that we can use Eq. 18-10 to determine the current per unit area. Both currents are in the same direction in terms of conventional current – positive charge moving north has the same effect as negative charge moving south – and so they can be added. I neAvd I A nevd He nevd O 2.8 1012 ions m3 2 1.60 1019 C ion 2.0 10 6 m s 7.0 1011 ions m3 1.60 1019 C ion 7.2 106 m s 2.6 A m 2 , North 52. 53. The magnitude of the electric field is the voltage change per unit meter. V 70 103 V E 7.0 106 V m 8 x 1.0 10 m The speed is the change in position per unit time. x 7.20 102 m 3.40 102 m v 35 m s t 0.0063s 0.0052s Two measurements are needed because there may be a time delay from the stimulation of the nerve to the generation of the action potential. 54. The energy required to transmit one pulse is equivalent to the energy stored by charging the axon capacitance to full voltage. In Example 18-15, the capacitance is estimated at 1.0 10 8 F , and the potential difference is about 0.1 V. E 12 CV 2 0.5 108 F 0.1V 5 1011 J 2 The power is the energy per unit time, for 10,000 neutrons transmitting 100 pulses each per second. P E t 5 10 11 J neuron 1.0 104 neurons 0.01s 5 10 5 W 55. The power is the work done per unit time. The work done to move a charge through a potential difference is the charge times the potential difference. The charge density must be multiplied by the surface area of the cell (the surface area of an open tube, length times circumference) to find the actual charge moved. P W QV Q V t mol ions 19 C 6 3 107 2 6.02 1023 1.6 10 0.10 m 20 10 m 0.030 V m s mol ion t t 5.4 10 9 W © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 56 Physics: Principles with Applications, 6th Edition Giancoli 56. 1C s 3600 s 3600 C 1A 1 h 1.00A h 57. Use Eq. 18-5 to calculate the current. P 746 W P IV I 6.22 A V 120 V 58. The energy supplied by the battery is the energy consumed by the lights. Esupplied Econsumed QV Pt t QV P 95A h 3600 s h 12V 92 W 1h 12.39 h 12 h 3600 s 44609 s 59. Use Eq. 18-6 to express the resistance in terms of the power, and Eq. 18-3 to express the resistance in terms of the wire geometry. P 4 V2 R L d 2 R V2 P V2 P d R 4 LP V 2 L A L r 2 4 L d2 4 9.71 108 m 5.4 m 1500 W 110 V 2.9 104 m 2 60. From Eq. 18-2a, if R V I , then G I V I 0.73 A G 0.24S V 3.0 V 61. To deliver 10 MW of power at 120 V requires a current of I P V 107 W 120 V 83, 300 A . Calculate the power dissipated in the resistors using the current and the resistance. 2 1.0 m L L L 2 P I 2 R I 2 I 2 2 4I 2 4 83, 300 A 1.68 10 8 m 2 A r d 5.0 103 m 2 1.187 107 W 1kW $0.10 1h 1000 W kWh Cost Power time rate per kWh 1.187 107 W $1187 $1, 200 per hour per meter 62. (a) Calculate the total kWh used per day, and then multiply by the number of days and the cost per kWh. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 57 Physics: Principles with Applications, 6th Edition Giancoli 1.8 kW 3.0 h d 4 0.1kW 6.0 h d 3.0 kW 1.4 h d 2.0 kWh d 14.0 kWh d $0.105 $44.10 per month kWh Cost 14.0 kWh d 30 d (b) The energy required by the household is 35% of the energy that needs to be supplied by the power plant. Household Energy 0.35 coal mass coal energy per mass coal mass Household Energy 0.35 coal energy per mass 14.0 kWh d 365 d 1000 W 3600 s kW 1h kcal 4186 J 0.35 7000 kg 1kcal 1794 kg 1800 kg of coal 63. the The length of the new wire is half the length of the original wire, and the cross-sectional area of new wire is twice that of the original wire. Use Eq. 18-3. 1 L L L L R0 0 L 12 L0 A 2 A0 R 2 0 14 0 14 R0 A0 A 2 A0 A0 The new resistance is one-fourth of the original resistance. 64. (a) Use Eq. 18-6 to relate the power to the voltage for a constant resistance. P V2 R P105 P117 105 V 2 117 V 2 105 V 2 R 117 V 2 R 0.805 or a 19.5% decrease (b) The lower power output means that the resistor is generating less heat, and so the resistor’s temperature would be lower. The lower temperature results in a lower value of the resistance, which would increase the power output at the lower voltages. Thus the decrease would be somewhat smaller than the value given in the first part of the problem. 65. The Assume that we have a meter of wire, carrying 35 A of current, and dissipating 1.8 W of heat. power dissipated is PR I 2 R , and the resistance is R PR I 2 R I 2 L A I2 A . L 4 L I2 2 r d2 L d I 2I 2 35 A PR PR 2 L 4 L 1.68 10 8 m 1.0 m 1.8 W 3.8 103 m © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 58 Physics: Principles with Applications, 6th Edition Giancoli 66. (a) (b) (c) The angular frequency is 210 rad s . 210 rad s f 33.4 Hz 2 2 The maximum current is 1.80 A. I 1.80 A I rms max 1.27 A 2 2 For a resistor, V IR . V IR 1.80 A sin 210 t 42.0 75.6sin 210 t V 67. (a) (b) The power delivered to the interior is 65% of the power drawn from the source. P 950 W Pinterior 0.65 Psource Psource interior 1462 W 1500 W 0.65 0.65 The current drawn is current from the source, and so the source power is used to calculate the current. Psource IVsource I 68. Psource Vsource 1462 W 120 V 12.18 A 12 A The volume of wire is unchanged by the stretching. The volume is equal to the length of the wire times its cross-sectional area, and so since the length was increased by a factor of 3, the area was decreased by a factor of 3. Use Eq. 18-3. L 3L L L R0 0 L 3L0 A 13 A0 R 1 0 9 0 9 R0 9.00 A0 A A0 A0 3 69. The long, thick conductor is labeled as conductor number 1, and the short, thin conductor is labeled V2 as number 2. The power transformed by a resistor is given by Eq. 18-6, P ,and both have the R same voltage applied. L L R1 1 R2 2 L1 2 L2 A1 4 A2 diameter1 2diameter2 A1 A2 P1 P2 2 V1 R1 2 2 V R2 R2 R1 L2 A2 L2 A1 1 4 2 L1 L1 A2 2 A1 P1 : P2 2 :1 70. The heater must heat 124 m3 of air per hour from 5o C to 20 o C , and also replace the heat being lost at a rate of 850 kcal/h. Use Eq. 14-2 to calculate the energy needed to heat the air. The density of air is found in Table 10-1. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 59 Physics: Principles with Applications, 6th Edition Giancoli Q mcT Q t Power required 408 71. (a) m3 kg kcal kcal 15Co 408 1.29 3 0.17 o h m kg C h kcal h 850 kcal h 1258 kcal 4186 J 1h 1463 W 1500 W h kcal 3600s V2 11cents kWh R V2 mcT 0.75 Poven 240 V 2 26.18 26 R P 2200 W Only 75% of the heat from the oven is used to heat the water. 0.75 Poven t Heat absorbed by water mcT t (c) t cT 124 Use Eq. 18-6. P (b) m 1kg o o 4186 J kg C 85 C 1L 25.88s 26 s 0.120 L 0.75 2200 W 2.2 kW 25.88s 1h 3600 s 0.17 cents 72. (a) The horsepower required is the power dissipated by the frictional force, since we are neglecting the energy used for acceleration. 1m s 1hp P Fv 240 N 45 km hr 3000W 4.0 hp 3.6 km hr 746 W (b) The charge available by each battery is Q 95A h 95C s 3600s 3.42 105 C , and so the total charge available is 24 times that. The potential energy of that charge is the charge times the voltage. That energy must be delivered (batteries discharged) in a certain amount of time to produce the 3000 W necessary. The speed of the car times the discharge time is the range of the car between recharges. PE QV QV d P t t t P v d vt v 73. the QV P v QV Fv QV F 24 3.42 105 C 12 V 240 N 410 km The mass of the wire is the density of copper times the volume of the wire, and the resistance of wire is given by Eq. 18-3. We represent the mass density by m and the resistivity by . © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 60 Physics: Principles with Applications, 6th Edition Giancoli R L L mR A 74. 75. A m L R L m m LA m L R L R 0.018 kg 12.5 8.9 10 d 1 2 3 kg m3 1.68 10 8 m d 2 4 L R 38.79 m 38.8 m 4 1.68 10 8 m 38.79 m 12.5 2.58 104 m Use Eq. 18-6. (a) P (b) P V2 R V2 R 120 V 2 12 120 V 2 140 1200 W 103 W 100 W Model the protons as moving in a continuous beam of cross-sectional area A. Then by Eq. 18-10, N , where N is the I neAvd . The variable n is the number of protons per unit volume, so n Al number of protons in the beam and l is the circumference of the ring. The “drift” velocity in this case is the speed of light. N N I neAvd eAvd evd Al l 3 11 10 A 6300 m Il N 1.4 1012 protons evd 1.60 10 19 C 3.00 108 m s 76. A The power is given by P IV . (a) P IV 12 A 220 V 2640W 2600 W The power dissipated is given by PR I 2 R , and the resistance is R (b) PR I R I 2 2 L A . 8 L 2 4 1.68 10 m 15 m 2 4 L I I 12 A 17.433 W 2 A r2 d2 1.628 103 m L 2 17 W 4 L 12 A 4 1.68 108 m 15 m (c) PR I 2 (d) The savings is due to the power difference. d2 2 2.053 10 m 3 2 10.962 W 11W 1kW 12 h $0.12 30 d 1000 W 1d 1kWh Savings 17.433 W 10.962 W $0.2795 / month 28 cents per month © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 61 Physics: Principles with Applications, 6th Edition Giancoli 77. The resistance can be calculated from the power and voltage, and then the diameter of the wire can be calculated from the resistance. V2 V2 L L V2 L P R R 2 2 R P A 12 d P 12 d 4 LP d 78. 79. V 2 4 100 10 8 m 3.8 m 95 W 120 V 2 1.8 10 4 m Use Eq. 18-6 for the power in each case, assuming the resistance is constant. V2 R P13.8 V 13.82 13.8 V 2 1.3225 32% increase P12.0 V 12.02 V R (a) 12.0 V The current can be found from Eq. 18-5. I PV (b) I A PA VA 40W 120 V 0.33 A The resistance can be found from Eq. 18-6. R (c) I B PB VB 40W 12 V 3.3 A 120 V 2 VA2 V2 RA 360 P PA 40 W The charge is the current times the time. RB VB2 PB 12 V 2 40 W 3.6 QA I At 0.33 A 3600 s 1200 C Q It QB I B t 3.3 A 3600 s 12, 000 C (d) The energy is the power times the time, and the power is the same for both bulbs. E A EB 40 W 3600 s 1.44 105 J E Pt (e) Bulb B requires a larger current, and so should have larger diameter connecting wires to avoid overheating the connecting wires. 80. The current in the wire can be found by I P V . (a) PR I 2 R (b) PR P2 V2 R 2250 W 2 120 V P2 4 L V d 2 2 2 P2 L P2 L P2 4 L V 2 A V 2 r2 V 2 d 2 4 1.68 10 8 m 25.0 m 28.0 W 2 2.59 103 m 2250 W 2 120 V 2 4 1.68 108 m 25.0 m 4.12 103 m 2 11.1W 81. Eq. 18-3 can be used. The area to be used is the cross-sectional area of the pipe. 1.68 10 8 m 10.0 m L L R 1.34 104 2 2 2 2 2 2 A r r 2.50 10 m 1.50 10 m outside inside © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 62 Physics: Principles with Applications, 6th Edition Giancoli 82. The volume of wire Volume length cross-sectional area remains constant as the wire is stretched from the original length of L0 to the final length of 2L0 . Thus the cross-sectional area 1 2 changes from A0 to R0 P0 L0 A0 V2 A0 . Use Eq. 18-3 for the resistance and 18-6 for the power dissipated. R P L V2 A 2 L0 1 2 V2 A0 1 4 4 V2 R0 R 4 R0 R0 The power is reduced by a factor of 4 . L0 A0 4 R0 14 P0 83. The resistance of the filament when the flashlight is on is R V 3.2 V I 0.20 A with a combination of Eqs. 18-3 and 18-4 to find the temperature. R R0 1 T T0 T T0 16 . That can be used 1 R 1 16 o 1 2168o C 2200o C 1 20 C 1 o R0 0.0045 C 1.5 © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 63