Physics Solution Manual for 1100 and 2101

7 since the voltage and the current are known we can

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Unformatted text preview: __________________________________________________________________ 81. REASONING AND SOLUTION Consider one revolution of either rod. The magnitude ξ of the emf induced across the rod is () B π L2 ∆A ξ = −B = ∆t ∆t The angular speed of the rods is ω = 2π /∆t, so ξ = 1 BL2ω. The rod tips have opposite 2 polarity since they are rotating in opposite directions. Hence, the difference in potentials of the tips is 2 ∆V = BL ω so ∆V 4.5 × 103 V = = 2100 rad/s BL2 ( 4.7 T )( 0.68 m )2 ______________________________________________________________________________ ω= 82. REASONING AND SOLUTION a. On startup, the back emf of the generator is zero. Then, R= V 117 V = = 9.59 Ω I 12.2 A b. At normal speed ξ = V – IR = 117 V – (2.30 A)(9.59 Ω) = 95 V Chapter 22 Problems c. The back emf of the motor is proportional to the rotational speed, so at 1 3 1239 the normal speed, the back emf is 1 3 (95 V) = 32 V The voltage applied to the resistor is then V = 117 V – 32 V = 85 V, so the current is V 85 V = = 8.9 A R 9.59 Ω ______________________________________________________________________________ I= CHAPTER 23 ALTERNATING CURRENT CIRCUITS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. 2 (d) According to P = Vrms / R (Equation 20.15c), the average power is proportional to the square of the rms voltage. Tripling the voltage causes the power to increase by a factor of 2 3 = 9. 2. Irms = 1.9 A 3. (b) The current Irms through a capacitor depends inversely on the capacitive reactance XC, as expressed by the relation Irms = Vrms/ XC (Equation 23.1). The capacitive reactance becomes infinitely large as the frequency goes to zero (see Equation 23.2), so the current goes to zero. 4. (e) According to X C = 1/ ( 2π f C ) (Equation 23.2) and X L = 2π f L (Equation 23.4), doubling the frequency f causes XC to decrease by a factor of 2 and XL to increase by a factor of 2. 5. Irms = 1.3 A 6. (a) The component of the phasor along the vertical axis is V0 sin 2π f t (see the drawing that accompanies this problem), which is the instantaneous value of the voltage. 7. (b) The instantaneous value of the voltage is the component of the phasor that lies along the vertical axis (see Sections 23.1 and 23.2). This vertical component is greatest in B and least in A, so the ranking is (largest to smallest) B, C, A. 8. (d) In a resistor the voltage and current are in phase. This means that the two phasors are colinear. 9. (c) Power is dissipated by the resistor, as discussed in Section 20.5. On the other hand, the average power dissipated by a capacitor is zero (see Section 23.1). 10. Irms = 2.00 A 11. (a) When the rms voltage across the inductor is greater than that across the capacitor, the voltage across the RCL combination leads the current (see Section 23.3). Chapter 23 Answers to Focus on Concepts Questions 1241 12. (d) Since Irms= Vrms/Z (Equation 23.6), the current is a maximum when the impedance Z is a minimum. The impedance is Z = R 2 + ( X L − X C ) 2 (Equation 23.7), and it has a minimum value when XC = XL = 50 Ω. 13. (c) The inductor has a very small reactance at low frequencies and behaves as if it were replaced by a wire with no resistance. Therefore, the circuit behaves as two resistors, R1 and R2, connected in parallel. The inductor has a very large reactance at high frequencies and behaves as if it were cut out of the circuit, leaving a gap in the connecting wires. The circuit behaves as a single resistance R2 connected across the generator. The situation at low frequency gives rise to the largest possible current, because the effective resistance of the parallel combination is smaller than the resistance R2. 14. (a) The capacitor has a very small reactance at high frequencies and behaves as if it were replaced by a wire with no resistance. Therefore, the circuit behaves as two resistors, R1 and R2, connected in parallel. The capacitor has a very large reactance at low frequencies and behaves as if it were cut out of the circuit, leaving a gap in the connecting wires. Therefore, the circuit behaves as a single resistor R1 connected across the generator. The situation at high frequencies gives rise to the largest possible current, because the effective resistance of the parallel combination is smaller than the resistance R1. 15. (e) At low frequencies, the capacitor has a very large reactance. In the series circuit, this large reactance gives rise to a large impedance and, hence, a small current. The parallel circuit has the larger current, because current can flow through the inductor, which has a small reactance at low frequencies. 16. (b) The resonant frequency f0 is given by f0 = 1 2π LC (Equation 23.10). It depends only on C and L, and not on R. 17. f0 = 1.3 × 103 Hz 18. (d) The resonant frequency f0 is given by f0 = 1 (Equation 23.10). When a second 2π LC capacitor is added in parallel, the equivalent capacitance increases (see Section 20.12). Therefore, the resonant frequency decreases. 1242 ALTERNATING CURRENT CIRCUITS...
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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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