Chapter 24: Alternating-Current CircuitsAnswers to Even-Numbered Conceptual Questions2. Current and voltage are not always in phase in an ac circuit because capacitors and inductors respond not tothe current itself—as a resistor does—but to the charge (capacitor) or to the rate of change of the current(inductor). The charge takes time to build up; therefore, a capacitor’s voltage lags behind the current. Therate of change of current is greatest when the current is least; therefore, an inductor’s voltage leads thecurrent. Resistors, of course, are always in phase with the current.4. As the frequency is increased, the inductive reactance increases as well. Therefore, at frequencies greaterthan the resonance frequency of an LC circuit, the inductive reactance is greater than the capacitivereactance. As a result, the inductor dominates, and the voltage leads the current. This means that the phaseangle, , is positive.6. No. In a dc circuit the frequency is zero, which means that the inductive reactance is zero as well. Therefore,the inductor has no effect at all on the current in the circuit, nor on the brightness of the bulb. In a dc circuit,an ideal inductor is the same as a piece of zero-resistance wire.8. Recall that charge in an RLC circuit is the analog of position in a mass-spring system. Therefore, the current,which is the rate of change of charge, is analogous to the velocity, which is the rate of change of position.(See Table 24-2.)10. Yes. All that is required for their resonance frequencies to be the same is for the product of L and C to be thesame. (See equation 24-18.)Solutions to Problems and Conceptual Exercises1. Picture the Problem: The voltage in an ac generator oscillates between −55 V and +55 V.Strategy: Divide the peak voltage by the square root of two to calculate the rms voltage. Solution: Calculate the rms voltage: Vrms=12Vmax=1255 V = 39 VInsight: The root-mean-square voltage is about 70% of the peak voltage.2. Picture the Problem: The voltage in the European wall socket oscillates between the positive and negative peak voltages, resulting in an rms voltage of 240 V.Strategy: Multiply the rms voltage by the square root of two to calculate the peak voltage. Solution: Calculate the peak voltage: Vmax= 2Vrms= 2 240 V( )= 340 VInsight: The European wall socket oscillates between voltages of −340 V and +340 V.3. Picture the Problem: The peak current will flow through a resistive circuit when the voltage is maximum.Strategy: Multiply the rms voltage by the square root of two to calculate the peak voltage. Then use Ohm’s Law (equation 21-2) to calculate the resistance in the circuit.Solution: 1. Calculate the peak voltage: Vmax= 2Vrms= 2 120( )=170 VCopyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.24 – 1Chapter 24: Alternating-Current Circuits James S. Walker, Physics, 4th Edition2. Solve Ohm’s Law for the resistance: R =VmaxImax=170 V2.1 A= 81 ΩInsight: An alternative method of solving this problem would be to convert the peak current to the rms current (1.48 A).Then divide the rms voltage by the rms current to obtain the resistance.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.24 – 2Chapter 24: Alternating-Current Circuits James S. Walker, Physics, 4th Edition4. Picture the Problem: Power is dissipated at a rate proportional to the square of the current in a resistive AC circuit.Strategy: Solve equation 21-5 for the average power dissipated, using the rms current. To calculate the peak power dissipated, replace the rms current with the peak current, using equation 24-3. Solution: 1. (a) Use equation 21-5 to find Pav: Pav=Irms2R = 0.85 A( )2150 Ω( )=110 W = 0.11 kW2. (b) Calculate the peak power dissipated: Pmax=Imax2R = 2Irms( )2R =2 Irms2R( )=2 110 W( )= 0.22 kWInsight: When the current oscillates as a sine wave, the average power dissipation is one-half the peak power dissipation.5. Picture the Problem: Power is dissipated at a rate proportional to the square of the voltage in a resistive AC circuit.Strategy: Solve equation 24-6 for the average power dissipation and equation 21-6 for the peak power dissipation. Solution: 1. (a) Use equation 24-6 to find Pav: Pav=12Vmax2R=12141 V( )23.33×103Ω( )= 2.99 W2. (b) Calculate the peak power dissipation: Pmax=Vmax2R=141 V( )23.33×103Ω= 5.97 WInsight: For a sinusoidal voltage the peak power dissipation in a resistor is double the average power dissipation.6. Picture the Problem: A light bulb dissipates power as the voltage oscillates across its filament resistance.Strategy: Calculate the resistance from the average power and the rms voltage using equation 21-6. Then, from the resistance and rms voltage, solve for the rms current using Ohm’s Law (equation 21-2). Convert the rms current to maximum current by multiplying it by the square root of two. Finally, use the resistance and maximum current to calculate the peak power dissipation.Solution: 1. (a) Solve equation 21-6 for R: R =Vrms2Pav=120 V( )275 W= 190 Ω2. (b) Use Ohm’s Law to calculate Irms: Irms=VrmsR=120 V192 Ω=0.625 A3. Convert to peak current: Imax= 2Irms= 2 0.625 A( )= 0.88 A4. (c) Calculate maximum power: Pmax=Imax2R = 0.884 A( )2192 Ω( )=150 W = 0.15 kWInsight: The peak power dissipated is equal to twice the average power dissipated.7. Picture the Problem: The figure shows a voltage oscillating as a square wave between −5.0 V and +5.0 V. The rms voltage in this case is not equal to the peak voltage divided by the square root of two.Strategy: To calculate the rms voltage, first square the voltages at each time period. Average the result and take the square root.Solution: 1. Calculate V 2 when V = +5.0 V: V+2= 5.0 V( )2=25 V2Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.24 – 3Chapter 24: Alternating-Current Circuits