Physics Solution Manual for 1100 and 2101

# Thus the current in the circuit obeys the relation it

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Unformatted text preview: CHAPTER 23 ALTERNATING CURRENT CIRCUITS PROBLEMS 1. SSM REASONING The voltage across the capacitor reaches its maximum instantaneous value when the generator voltage reaches its maximum instantaneous value. The maximum value of the capacitor voltage first occurs one-fourth of the way, or one-quarter of a period, through a complete cycle (see the voltage curve in Figure 23.4). SOLUTION The period of the generator is T = 1/ f = 1/ (5.00 Hz) = 0.200 s . Therefore, the least amount of time that passes before the instantaneous voltage across the capacitor 1 1 reaches its maximum value is 4 T = 4 ( 0.200 s) = 5.00 × 10 –2 s . 2. REASONING When two capacitors of capacitance C are connected in parallel, their equivalent capacitance CP is given by CP = C + C = 2C (Equation 20.18). We will find the 1 (Equation 23.2). 2π f C P Because the capacitors are the only devices connected to the generator, the rms voltage Vrms across the capacitors is equal to the rms voltage of the generator. Therefore, the capacitive reactance XC is related to the rms voltage of the generator and the rms current Irms in the capacitive reactance XC of the equivalent capacitance from X C = circuit by Vrms = I rms X C (Equation 23.1). SOLUTION Substituting CP = 2C into X C = 1 (Equation 23.2) and solving for C, 2π f C P we obtain XC = 1 1 1 = = ( 2C ) 4π fC 2π f CP 2π f or Solving Vrms = I rms X C (Equation 23.1) for XC yields X C = C= Vrms I rms 1 4π fX C . Substituting this result into Equation (1), we obtain C= 1 = 4π fX C (1) I rms 1 0.16 A = = = 8.7 ×10−7 F Vrms 4π fVrms 4π ( 610 Hz ) ( 24 V ) 4π f I rms Chapter 23 Problems 3. 1243 SSM REASONING The rms current in a capacitor is Irms = Vrms/XC, according to Equation 23.1. The capacitive reactance is XC = 1/(2π f C), according to Equation 23.2. For the first capacitor, we use C = C1 in these expressions. For the two capacitors in parallel, we use C = CP, where CP is the equivalent capacitance from Equation 20.18 (CP = C1 + C2). Taking the difference between the currents and using the given data, we can obtain the desired value for C2. The capacitance C1 is unknown, but it will be eliminated algebraically from the calculation. SOLUTION Using Equations 23.1 and 23.2, we find that the current in a capacitor is I rms = V rms XC = V rms b 1 / 2π f C = gV rms 2π f C Applying this result to the first capacitor and the parallel combination of the two capacitors, we obtain I 1 = V rms 2 π f C1 and I Combination = V rms 2 π f C1 + C 2 1442443 1444442444443 c Single capacitor h Parallel combination Subtracting I1 from ICombination, reveals that c h I Combination − I 1 = V rms 2 π f C1 + C2 − V rms 2 π f C1 = V rms 2 π f C2 Solving for C2 gives C2 = 4. I Combination − I 1 V rms 2 π f = 0.18 A = 2 .7 × 10 −6 F 24 V 2 π 440 Hz b gb g REASONING As the frequency f of the generator increases, the capacitive reactance XC of 1 the capacitor decreases, according to X C = (Equation 23.2), where C is the 2π f C capacitance of the capacitor. The decreasing capacitive reactance leads to an increasing rms V current Irms, as we see from I rms = rms (Equation 23.1), where Vrms is the constant rms XC voltage across the capacitor. We know that Vrms is constant, because it is equal to the constant rms generator voltage. The fuse is connected in series with the capacitor, so both have the same current Irms. We will use Equations 23.1 and 23.2 to determine the frequency f at which the rms current is 15.0 A. 1244 ALTERNATING CURRENT CIRCUITS SOLUTION Solving Equation 23.2 for f, we obtain f= 1 2π C X C (1) In terms of the rms current and voltage, Equation 23.1 gives the capacitive reactance as V X C = rms . Substituting this relation into Equation (1) yields I rms f= 5. 1 = 2π C X C 1 V 2π C rms I rms = I rms 2π CVrms = ( 15.0 A ) 2π 63.0 × 10 −6 F ( 4.00 V ) = 9470 Hz 1 (Equation 23.2), 2π f C where f is the frequency of the ac current and C is the capacitance of the capacitor. We note that XC is inversely proportional to f for a given value of C. Therefore, we will be able to solve this problem by applying Equation 23.2 twice, once for each value of the frequency and each time with the same value of the capacitance. REASONING The reactance XC of a capacitor is given as X C = SOLUTION Applying Equation 23.2 for each value of the frequency, we obtain X C, 870 = 1 2π f870 C and X C, 460 = 1 2π f 460 C Dividing the equation on the left by the equation on the right and noting that the unknown capacitance C is eliminated algebraically, we find that X C, 870 X C, 460 1 f 2π f870 C = = 460 1 f870 2π f 460 C Solving for the reactance at a frequency of 870 Hz gives X C, 870 = X C, 460 f 460 f870 460 Hz = ( 68 Ω ) = 36 Ω 870 Hz Chapter 23 Problems 6. 1245 REASONING The rms current in the circuit is given by Irms = Vrms/XC (Equation 23.1), where Vrms is the rms voltage of the generator, and XC = 1/(2π f C) is the capacitive reactance (see Equation 23.2). For a given voltage, smaller reactances lead to greater currents. W...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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