CHAPTER 23ALTERNATING CURRENTCIRCUITSCONCEPTUAL QUESTIONS_____________________________________________________________________________________________1. REASONING AND SOLUTION A light bulb and a parallel plate capacitor (including a dielectricmaterial between the plates) are connected in series to the 60-Hz ac voltage present at a wall outlet.Since the capacitor and the light bulb are in series, the rms current at any instant is the same througheach element, and is given by Equation 23.6: Irms=Vrms/ Z, where Z is the impedance of the circuit.The impedance of the circuit is given by Equation 23.7 with XL = 0 (since there is no inductance inthe circuit): Z=R2+(– XC)2=R2+XC2. According to Equation 19.10, if the dielectricbetween the plates of the capacitor is removed, the capacitance decreases by a factor of κ, where κ isthe dielectric constant. From Equation 23.2, XC=1 / (2πfC ), we see that decreasing the capacitanceincreases the capacitive reactance XC. Therefore, the impedance of the circuit, R2+XC2,increases. The rms current in the circuit is Irms=Vrms/ Z and will, therefore, be less than it wasbefore the dielectric was removed. As a result, the brightness of the bulb will decrease._____________________________________________________________________________________________2. REASONING AND SOLUTION The ends of a long straight wire are connected to the terminals ofan ac generator, and the current is measured. The wire is then disconnected, wound into the shape ofa multiple-turn coil, and reconnected to the generator. After the wire is wound into a coil, it has agreater inductance. When the generator is turned on, the coil will develop a voltage that opposes achange in the current according to Faraday's law. Since the induced voltage opposes the rise incurrent, the rms current in the circuit will be less than it was before the wire was wound into a coil.This can also be seen by considering Equations 23.6 and 23.7. Before the wire was wound into acoil, its primary property was that of resistance, and the current through the wire was given byIrms=Vrms/ Z with Z = R, or Irms=Vrms/ R. After the wire is wound into a coil, the wire possessesboth resistance and inductance. Now the current in the coil is given by Irms=Vrms/ Z, whereZ=R2+XL2. Since Z is necessarily greater than R, the rms current is less when the wire iswound into a coil._____________________________________________________________________________________________3. REASONING AND SOLUTION An air-core inductor is connected in series with a light bulb ofresistance R. This circuit is plugged into an electrical outlet. The current in the circuit is given byEquation 23.6, Irms=Vrms/ Z, where, from Equation 23.7, Z=R2+XL2. When a piece of ironis inserted in the inductor, the magnetic field in the inductor is enhanced relative to that in air, and theinductance increases. Equation 23.4, XL=2πfL, shows that when the inductance L increases, theinductive reactance XL also increases. The impedance of the circuit, therefore, increases, and thecurrent in the circuit, Irms=Vrms/ Z, decreases. Thus, the brightness of the bulb decreases._____________________________________________________________________________________________Chapter 23 Conceptual Questions 1334. REASONING AND SOLUTION Consider two ac circuits. The generators in each circuit areidentical (same frequency, same rms voltage). In circuit I, only a resistor is connected across thegenerator. In circuit II, the same resistor is in series with an inductor.The average power used by an ac circuit is given by Equation 23.9: P=IrmsVrms cos φ, where theterm cos φ is the power factor of the circuit and is equal to R/Z. Therefore, P=IrmsVrmsR / Z. Butaccording to Equation 23.6, Irms=Vrms/ Z. Therefore,P=VrmsZ      VrmsRZ=Vrms2RZ2.For circuit I, the impedance is Z = R, so that PI=Vrms2/ R. For circuit II, Equation 23.7 indicates thatZ2=R2+XL2. Substituting this expression for Z2 we find thatPII=Vrms2RR2+XL2=Vrms2/ R1+XL2/ R2=PI1+XL2/ R2.Thus, since the term XL2/ R2 in the denominator on the right is positive, we find that PI is greaterthan PII. Therefore, circuit I consumes more average power._____________________________________________________________________________________________5. REASONING AND SOLUTION Consider the series RCL circuit shown in Figure 23.9. Theimpedance of the circuit is given by Equation 23.7: Z=R2+(XL−XC)2, where the capacitivereactance is given by Equation 23.2 [XC=1/(2πfC )] and the inductive reactance is given byEquation 23.4 [XL=2πfL]. One example of this expression for Z is plotted in Figure 23.10 (see thered curve). The vertical axis in this figure gives the impedance, while the horizontal axis gives thefrequency. A horizontal line drawn to intersect the vertical axis above the minimum in the curve willintersect the curve at two places. These places correspond to two different frequencies on thehorizontal axis. Thus, the circuit in Figure 23.9 can have the same impedance at two differentfrequencies._____________________________________________________________________________________________6. REASONING AND SOLUTION An inductor and a capacitor are connected in parallel across theterminals of a generator.a. The frequency of the generator becomes very large. From Equation 23.2 (or Figure 23.2) we seethat, in the high frequency limit, the capacitive reactance approaches zero. Therefore, in the highfrequency limit, the capacitor behaves as if it were replaced by a wire with zero resistance, and thegenerator delivers a very large current to the capacitor. From Equation 23.4 (or Figure 23.6) we seethat, in the high frequency limit, the inductive reactance becomes very large. In this limit, theinductor behaves as if it has been cut out of the circuit, leaving a gap in the wire, and the generatordelivers no current to the inductor. Therefore, when the frequency of the generator becomes verylarge, the current through the capacitor becomes large, and no current flows through the inductor.The total current delivered by the generator is large.134 ALTERNATING CURRENT CIRCUITSb. The frequency of the generator becomes very small. From Equation 23.2 we see that, in the lowfrequency limit, the capacitive reactance approaches infinity. The capacitor acts as if it has been cutfrom the circuit, leaving a gap in the wire. As a result, the generator delivers no current to thecapacitor.