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Unformatted text preview: C HAPTER 21 Quick Quizzes 1. (c). The average power is proportional to the rms current which is nonzero 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 C H A P T E R 2 1 Problem Solutions 21.1 (a) ( 29 ( 29 max rms 2 2 100 V V V ∆ = ∆ = = 141 V (b) rms rms 100 V 5.00 V I R ∆ = = = Ω 2 0.0 A (c) max max 141 V 5.00 V I R ∆ = = = Ω 2 8.3 A or ( 29 max rms 2 2 20.0 A I I = = = 2 8.3 A (d) ( 29 ( 29 2 2 3 av rms 20.0 A 5.00 2.00 10 W I R = = Ω = = 2 .0 0 k W 21.2 ( 29 2 2 2 max 2 max max av rms 1 2 2 2 V I V I R R R R R ∆ ∆ = = = = , so ( 29 2 max av 2 V R ∆ = (a) If av 75.0 W = , then ( 29 ( 29 2 170 V 2 75.0 W R = = 193 Ω (b) If av 100 W = , then ( 29 ( 29 2 170 V 2 100 W R = = 145 Ω 21.3 The meters measure the rms values of potential difference and current. These are max rms 100 V 2 2 V V ∆ ∆ = = = 70.7 V , and rms rms 70.7 V 24.0 V I R ∆ = = = Ω 2 .95 A 21.4 All lamps are connected in parallel with the voltage source, so rms 120 V V ∆ = for each lamp. Also, the current is rms av rms I V = ∆ and the resistance is rms rms R V I = ∆ . 1, rms 2, rms 150 W 120 V I I = = = 1.25 A and 1 2 120 V 1.25 A R R = = = 96.0 Ω 3, rms 100 W 120 V I = = 0 .8 3 3 A and 3 120 V 0.833 A R = = 144 Ω 178 C H A P T E R 2 1 21.5 The total resistance (series connection) is 1 2 8.20 10.4 18.6 eq R R R = + = Ω + Ω = Ω , so the current in the circuit is rms rms 15.0 V15....
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This homework help was uploaded on 04/21/2008 for the course PHY 213 taught by Professor Cao during the Spring '08 term at Kentucky.
 Spring '08
 Cao
 Current, Power

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