Chapter 21 - CHAPTER 21 Quick Quizzes 1(c The average power is proportional to the rms current which is non-zero even though the average current is

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C HAPTER 21 Quick Quizzes 1. (c). The average power is proportional to the rms current which is non-zero even though the average current is zero. (a) is only valid for an open circuit. (b) and (d) can never be true because i av = 0 for AC currents. 2. (b). Choices (a) and (c) are incorrect because the unaligned sine curves in Figure 21.9 mean the voltages are out of phase, and so we cannot simply add the maximum (or rms) voltages across the elements. (In other words, V V R + V L + V C even though v = v R + v L + v C .) 3. (b). Note that this is a DC circuit. However, changing the amount of iron inside the solenoid changes the magnetic field strength in that region and results in a changing magnetic flux through the loops of the solenoid. This changing flux will generate a back emf that opposes the current in the circuit and decreases the brightness of the bulb. The effect will be present only while the rod is in motion. If the rod is held stationary at any position, the back emf will disappear and the bulb will return to its original brightness . 4. (b), (c). The radiation pressure (a) does not change because pressure is force per unit area. In (b), the smaller disk absorbs less radiation, resulting in a smaller force. For the same reason, the momentum in (c) is reduced. 177
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CHAPTER 21 Problem Solutions 21.1 (a) ( ) ( ) max rms 2 2 100 V VV ∆=∆ = = 141 V (b) rm s 100 V 5.00 V I R == = 20.0 A (c) V I R 28.3 A or ( ) 2 2 20.0 A II = == (d) () 2 23 av 20.0 A5 . 00 2.00 10 W IR ℘= = Ω= × = 2.00 kW 21.2 ( ) 2 2 2 2 1 22 2 V IV R R RR ⎛⎞ ⎡⎤ = = = ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ , so ( ) 2 2 V R = (a) If , then 75.0 W 2 170 V 2 75.0 W R = = 193 (b) If , then 100 W 2 2100 W R 145 21.3 The meters measure the rms values of potential difference and current. These are V V ∆= = = 70.7 V , and 24.0 V I R = 2.95 A 21.4 All lamps are connected in parallel with the voltage source, so 120 V V = for each lamp. Also, the current is = ℘∆ and the resistance is RVI = ∆ . 1,rm s 2,rm s 150 W = 1.25 A and 12 = 96.0 3,rm s I 0.833 A and 3 R = = 144 178
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CHAPTER 21 21.5 The total resistance (series connection) is 12 8.20 10.4 18.6 eq RRR = += + = , so the current in the circuit is rm s 15.0 V 0.806 A V I R == = The power to the speaker is then ( ) ( ) 2 2 av ker 0.806 A1 0 . 4 spea IR == = 6.76 W 21.6 (a) , so max 150 V V ∆= 22 V V = = 106 V (b) 377 rad s f ω ππ = 60.0 Hz (c) At () 1120 s t = , ( ) ( ) ( ) ( ) ( ) 150 Vs in 377 rad s 1 120 s1 5 0 i n rad v π ⎡⎤ ⎣⎦ = 0 (d) 50.0 V I R 3.00 A 21.7 1 2 C X fC = , so its units are 11V o l tV o l t O hm 1SecFarad 1Sec Coulom b Vol tC o u l ombSec Amp = = 21.8 ( ) rms 2 =2 = 2 C V II V X f C (a) ( ) ( ) ( ) 6 = 2 120 V2 6 0 .0 Hz 2.20 10 C/V =0.141 A I × = 141 mA (b) ( ) ( ) 6 = 2 240 V2 5 0 .0 Hz 2.20 C/V =0.235 A I × = 235 21.9 2 C V If C X V , so ( ) 6 0.30 A 2 24 . 01 0 F3 0 V I f CV = 2 4. 0 Hz × × 179
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CHAPTER 21 21.10 () max == 2 C V If C X π V ( ) ( ) ( ) 6 29 0 . 0 Hz 3.70 10 C /V 48.0 V=0 .100 A= 100 mA 21.11 rm s m ax 2 2 C VV Cf C V X ππ ∆∆ ⎛⎞ =∆ ⎜⎟ ⎝⎠ 2 , so 5 0.75 A 1. 71 0 F 26 0 H z 170 V2 I C fV = × = 17 F µ 21.12 2 C IC X ω , or ( ) 6 140 V = 120 rad s 6.00 F0 .
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This homework help was uploaded on 03/21/2008 for the course PY 211 / 212 taught by Professor Chilton during the Spring '08 term at N.C. State.

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Chapter 21 - CHAPTER 21 Quick Quizzes 1(c The average power is proportional to the rms current which is non-zero even though the average current is

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