Physics Solution Manual for 1100 and 2101

# 10 in order 2 lc to pick up the entire range of fm

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Unformatted text preview: /Z, where Vrms is the rms voltage of the generator and Z is the circuit impedance. The impedance of the circuit is given by Z = c R2 + X L − XC h (Equation 23.7). At 2 resonance the inductive reactance XL and the capacitive reactance XC are equal, so that Z = R. 2 SOLUTION Substituting Irms = Vrms/Z into P = I rms R , the average power delivered to the cricuit can be written as 2 P= 2 I rms R V = rms R Z (1) Substituting Z = R into Equation (1) yields 2 Vrms ( 3.0 V ) 2 Vrms Vrms P= = = 0.098 W R= R= Z R R 92 Ω 2 2 33. SSM REASONING The current in an RCL circuit is given by Equation 23.6, I rms = V rms / Z , where the impedance Z of the circuit is given by Equation 23.7 as Z= R 2 + ( X L − X C ) 2 . The current is a maximum when the impedance is a minimum for a given generator voltage. The minimum impedance occurs when the frequency is f 0 , corresponding to the condition that X L = X C , or 2 π f 0 L = 1 / ( 2 π f 0 C ) . Solving for the frequency f 0 , called the resonant frequency, we find that f0 = 1 2π LC Chapter 23 Problems 1263 Note that the resonant frequency depends on the inductance and the capacitance, but does not depend on the resistance. SOLUTION a. The frequency at which the current is a maximum is f0 = 1 2π LC = 1 2 π ( 17 .0 × 10 –3 H)( 12.0 × 10 –6 F) = 352 Hz b. The maximum value of the current occurs when f = f 0 . This occurs when X L = X C , so that Z = R . Therefore, according to Equation 23.6, we have I rms = 34. REASONING V rms Z = V rms R = 155 V = 15.5 A 10.0 Ω The resonant frequency of an RCL circuit is given by f0 = 1 2π LC (Equation 23.10), where L is inductance and C is the capacitance. Because only the inductance of this circuit changes, from L1 = 7.0 mH to L2 = 1.5 mH, we obtain the initial and final resonant frequencies from Equation 23.10: f 01 = 1 2π L1C and f 02 = 1 2π L2C (1) We will solve the first of Equations (1) for the capacitance C, and substitute the result into the second of Equations (1). SOLUTION Squaring both sides of the first of Equations (1) and solving for C, we obtain ( f01 )2 = 1 or ( 2π ) L1C 2 C= 1 (2) ( 2π f01 )2 L1 Substituting Equation (2) into the second of Equations (1) yields f 02 = 1 = 2π L2C 1 1 2π L2 2 ( 2π f 01 ) L1 = 2π f 01 2π L2 L1 = f 01 L1 L2 (3) In Equation (3) the initial resonant frequency is multiplied by the square root of the ratio of the inductances, so that if we express the initial resonant frequency in kHz, the final 1264 ALTERNATING CURRENT CIRCUITS resonant frequency will also be expressed in kHz, as requested. Similarly, we do not need to convert the inductances from millihenries to henries, since their units will cancel out. From Equation (3), then, the final resonant frequency of the circuit is f 02 = f 01 L1 L2 = (1.3 kHz ) 7.0 mH = 2.8 kHz 1.5 mH 35. REASONING As discussed in Section 23.4, a RCL series circuit is at resonance when the current is a maximum and the impedance Z of the circuit is a minimum. This happens when the inductive reactance XL equals the capacitive reactance XC. When XL = XC, the impedance of the circuit (see Equation 23.7) becomes Z = R 2 + ( X L − X C ) = R , so the 2 impedance is due solely to the resistance R. The average power dissipated in the circuit is 2 P = Vrms / R (Equation 20.15c). This relation can be used to find the power when the variable resistor is set to another value. SOLUTION The average power P dissipated when the resistance is R1 = 175 Ω is 1 2 P = Vrms / R1 . Likewise, the average power P2 dissipated when the resistance is R2 = 562 Ω 1 2 2 is P2 = Vrms / R2 . Solving the first equation for Vrms and substituting the result into the second equation gives P2 = 2 Vrms R2 = P R1 1 R2 = ( 2.6 W ) (175 Ω ) 562 Ω = 0.81 W d 36. REASONING The resonant frequency is given by Equation 23.10 as f 0 = 1 / 2 π LC i and is inversely proportional to the square root of the circuit capacitance C. Therefore, to reduce the resonant frequency, it is necessary to increase the circuit capacitance. The equivalent capacitance CP of two capacitors in parallel is CP = C1 + C2 (Equation 20.18), which is greater than either capacitance individually. Therefore, to increase the circuit capacitance, C2 should be added in parallel with C1. SOLUTION The initial resonant frequency is f01. The resonant frequency that results after C2 is added in parallel with C1 is f0P. Using Equation 23.10, we can express both of these frequencies as follows: f 01 = 1 2π LC1 and f0P = 1 2π LC P Chapter 23 Problems 1265 Here, CP is the equivalent parallel capacitance. Dividing the expression for f01 by the expression for f0P yields f 01 f 0P = d 1 / dπ 2 1 / 2π LC1 LC P i= i CP C1 According to Equation 20.18, the equivalent capacitance is CP = C1 + C2, so that this frequency ratio becomes f 01 f 0P = C1 + C 2 C1 = 1+ C2 C1 Squaring both sides of this result and solving for C2, we find 2 f01 C = 1+ 2 f C1 0P f 2 7.30 kHz 2 C2 = C1 01 − 1 = ( 2.60 µ F ) − 1 = 1.8 µ F f0P 5.60 kHz 37. REASONING Since the resonant frequency f0...
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