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# Anschp32

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1. 1. Chapter 32. Alternating-Current Circuits Physics, 6th Edition Chapter 32. Alternating-Current Circuits The Capacitor 32-1. A series dc circuit consists of a 4-µF capacitor, a 5000-Ω resistor, and a 12-V battery. What is the time constant for this circuit? τ = RC = (5000 Ω)(4 x 10-6 F) ; τ = 20 ms 32-2. What is the time constant for a series dc circuit containing a 6-µF capacitor and a 400-Ω resistor connected to a 20-V battery? τ = RC = (400 Ω)(6 x 10-6 F) ; τ = 2.40 ms 32-3. For the circuit described in Problem 32-1, what is the initial current and the final current? How much time is needed to be assured that the capacitor is fully charged? The initial current Io (before charge builds on capacitor) is determined by Ohm’s law: V 12 V I0 = = ; I0 = 2.40 mA R 5000 Ω It is assumed that a capacitor is fully charged after five time constants. t = 5τ = 5(20 ms); t = 100 ms The current is zero when the capacitor is fully charged; If = 0 A 32-4. For the circuit in Problem 32-2, what is the maximum charge for the capacitor and how much time is required to fully charge the capacitor? Qmax = CVB = (6 x 10-6 F)(20 V); Qmax = 120 µF t = 5RC = 5(2.40 ms); t = 12.0 ms 186
2. 2. Chapter 32. Alternating-Current Circuits Physics, 6th Edition *32-5. An 8-µF capacitor is connected in series with a 600-Ω resistor and a 24-V battery. After one time constant, what are the charge on the capacitor and the current in the circuit?  1 Q = CVB (1 − e −t RC ) = CVB  1 -  = CVB (0.632) ;  e Q = (8 µ F )(24 V)(0.632); Q = 121 µC VB −t RC 24 V  1  I= e =  ; I = 14.7 mA R 600 Ω  e  *32-6. Assume that the fully charged capacitor of Problem 32-5 is allowed to discharge. After one time constant, what are the current in the circuit and the charge on the capacitor? −VB −t RC −24 V  1  I= e =  ; I = -14.7 mA, decreasing R 600 Ω  e  1 Q = CVB e −t RC = (8 µ F)(24 V)   ; Q = 70.6 µC e *32-7. Assume that the 8-µF capacitor in Problem 32-5 is fully charged and is then allowed to decay. Find the time when the current in the circuit will have decayed to 20 percent of its initial value? [ Procedure: Let x = t/RC, solve for x, and then find t. ] −VB −t RC I 1 I= e = I max e − x ; = e− x ; e − x = 0.20; e+ x = = 5.0 R I max 0.2 t Take ln of both sides: x = ln (5.0) = 1.61 Now, x = = 1.61; t = 1.61RC RC t = 1.61(600 Ω)(24 V); t = 7.73 ms 187
3. 3. Chapter 32. Alternating-Current Circuits Physics, 6th Edition *32-8. A 5-µF capacitor is connected in series with a 12-V source of emf and a 4000 Ω resistor. How much time is required to place 40 µC of charge on the capacitor? [ Let x = t/RC ] Q = CVB (1 − e −t RC ); 40 µ C = (5 µ F)(12 V)(1 - e − x ); 0.667 = (1 - e − x ) 1 e − x = 1 − 0.667; e − x = 0.333; e+ x = = 3.0 0.333 t x = ln(3); = 1.10; t = 1.10 RC ; t = 22.2 ms RC The Inductor 32-9. A series dc circuit contains a 4-mH inductor and an 80-Ω resistor in series with a 12-V battery. What is the time constant for the circuit? What are the initial and final currents? L 0.004 H τ= = ; τ = 50.0 µs R 80 Ω VB V When t = 0, i= (1 − e − ( R / L )t ) = B (1 − 1) = 0 ; io = 0 R R VB V 12 V When t = ∞, i = (1 − e − ( R / L )t ) = B ( 1 − 0 ) ; i = ; if = 150 mA R R 80 Ω 32-10. A 5-mH inductor, a 160-Ω resistor and a 50-V battery are connected in series. What time is required for the current in the inductor to reach 63 percent of its steady state value? What is the current at that instant? (Current reaches 63% in one time constant.) L 0.005 H τ= = ; τ = 31.2 µs R 160 Ω 32-11. What is the current in Problem 32-9 after a time of one time constant? 188
4. 4. Chapter 32. Alternating-Current Circuits Physics, 6th Edition VB 12 V  1  i= (1 − e − ( R / L )t ) =  1 −  ; i = (0.15 A)(0.632); i = 94.8 mA R 80 Ω  e  32-12. Assume that the inductor in Problem 32-10 has reached its steady state value. If the circuit is broken how much time must pass before we can be sure the current in the inductor is zero? (After 5 time constants: t = 5τ) t = 5τ = 5(31.2 ms); t = 156 µs 32-13 The current in a 25-mH solenoid increases from 0 to 2 A in a time of 0.1 s. What is the magnitude of the self-induced emf? ∆i (2 A - 0) E = −L = −(0.025 H) ; E = - 500 mV ∆t 0.1 s 32-14. A series dc circuit contains a 0.05-H inductor and a 40-Ω resistor in series with a 90-V source of emf. What are the maximum current in the circuit and the time constant? VB 90 V imax = = ; imax = 2.25 A R 40 Ω L 0.05 H τ= = ; τ = 1.25 ms R 40 Ω *32-15. What is the instantaneous current in the inductor of Problem 32-14 after a decay time of 1.0 ms? (Assume the current decays from it’s maximum value of 2.25 A.)  40 Ω  1 1 Let x = (R/L)t: x =   (0.001 s); x = 0.80; e− x = x = 0.8 = 0.449  0.05 H  e e VB − ( R / L ) t VB − x 90 V i= e = e ; i= (0.449); i = 1.01 A R R 40 Ω You should show that the current after a rise time of 1 ms is 1.24 A. 189
5. 5. Chapter 32. Alternating-Current Circuits Physics, 6th Edition **32-16. A 6-mH inductor, a 50-Ω resistor, and a 38-V battery are connected in series. At what time after the connection will the current reach an instantaneous value of 600 mA? Procedure: Let x = (R/L)t:, then solve for x and substitute to find the time t: VB iR iR (0.6 A)(50 Ω) i= (1 − e − x ); = 1 − e− x ; e− x = 1 − = 1− ; e-x = 0.211 R VB VB 38 V 1 e+ x = = 4.75; Take ln of both sides: x = ln(4.75) = 1.56 0.211 Rt 1.56 L 1.56(0.006 H) x = 1.56 = ; t= = ; x = 187 µs L R 50 Ω Alternating Currents 32-17. An ac voltmeter, when placed across a 12-Ω resistor, reads 117 V. What are the maximum values for the voltage and current? Vmax = 2Veff = 1.414(117 V); Vmax = 165 V  117 V  I max = 2 I eff = 1.414   ; Imax = 13.8 A  12 Ω  32-18. In an ac circuit, the voltage and current reach maximum values of 120 V and 6.00 A. What are the effective ac values? I max 6A Vmax 120 V I eff = = ; Ieff = 4.24 A; Veff = = ; Veff = 84.9 V 2 1.414 2 1.414 32-19. A capacitor has a maximum voltage rating of 500 V. What is the highest effective ac voltage that can be supplied to it without breakdown? 190
6. 6. Chapter 32. Alternating-Current Circuits Physics, 6th Edition Vmax 500 V Veff = = ; Veff = 354 V 2 1.414 32-20. A certain appliance is supplied with an effective voltage of 220 V under an effective current of 20 A. What are the maximum and minimum values? Vmax = 2Veff = 1.414(220 V); Vmax = ±311 V I max = 2 I eff = 1.414(20 A); Imax = ±28.3 A Reactance 32-21. A 6-µF capacitor is connected to a 40-V, 60 Hz ac line. What is the reactance? What is the effective ac current in the circuit containing pure capacitance? 1 1 XC = = ; XC = 442 Ω 2π fC 2π (60 Hz)(6 x 10-6 F) Veff 40 V I eff = = ; Ieff = 90.5 mA XC 442 Ω 32-22. A 2-H inductor of negligible resistance is connected to a 50-V, 50 Hz ac line. What is the reactance? What is the effective ac current in the coil? X L = 2π fL = 2π (50 Hz)(2 H) = 628 Ω ; XL = 442 Ω Veff 50 V I eff = = ; Ieff = 79.6 mA XL 628 Ω 32-23. A 50-mH inductor of negligible resistance is connected to a 120-V, 60-Hz ac line. What is the inductive reactance? What is the effective ac current in the circuit? X L = 2π fL = 2π (60 Hz)(0.050 H) = 18.85 Ω; ; XL = 18.9 Ω 191
7. 7. Chapter 32. Alternating-Current Circuits Physics, 6th Edition Veff 120 V I eff = = ; Ieff = 6.37 mA XL 18.85 Ω 32-24. A 6-µF capacitor is connected to a 24-V, 50-Hz ac source. What is the current in the circuit? 1 1 XC = = ; XC = 531 Ω 2π fC 2π (50 Hz)(6 x 10-6 F) Veff 24 V I eff = = ; Ieff = 45.2 mA XC 531 Ω 32-25. A 3-µF capacitor connected to a 120-V, ac line draws an effective current of 0.5 A. What is the frequency of the source? Veff 120 V XC = = ; X C = 240 Ω I eff 0.5 A 1 1 1 XC = ; f = = ; f = 221 Hz 2π fC 2π CX C 2π (3 x 10 F)(240 Ω) -6 32-26. Find the reactance of a 60-µF capacitor in a 600-Hz ac circuit. What is the reactance if the frequency is reduced to 200 Hz? 1 1 XC = = ; XC = 4.42 Ω 2π fC 2π (600 Hz)(60 x 10-6 F) 1 1 XC = = ; XC = 13.3 Ω 2π fC 2π (200 Hz)(60 x 10-6 F) 32-27. The frequency of an alternating current is 200 Hz and the inductive reactance for a single inductor is 100 Ω. What is the inductance? XL 100 Ω X L = 2π fL; L= = ; L = 79.6 mH. 2π f 2π (200 Hz) 192
8. 8. Chapter 32. Alternating-Current Circuits Physics, 6th Edition 32-28. A 20-Ω resistor, a 2-µF capacitor, and a 0.70-H inductor are available. Each of these, in turn, is connected to a 120-V, 60 Hz ac source as the only circuit element. What is the effective ac current in each case? V 120 V Resistance only: i = = ; i = 6.00 A R 20 Ω 1 120 V Capacitor only: XC = = 1326 Ω; i = ; i = 90.5 mA 2π (60 Hz)(2 x 10 F) -6 1326 Ω 120 V Inductor only: X L = 2π fL = 2π (60 Hz)(0.7 H) = 264 Ω; i = ; i = 455 mA 264 Ω Series AC Circuits *32-29. A 300-Ω resistor, a 3-µF capacitor, and a 4-H inductor are connected in series with a 90- V, 50 Hz ac source. What is the net reactance of the circuit? What is the impedance? 1 XC = = 1061 Ω; X L = 2π (50 Hz)(4 H) = 1257 Ω; 2π (50 Hz)(3 x 10-6 F) Net Reactance: XL – XC = 196 Ω Z = R 2 + ( X L − X C ) 2 = (300 Ω) 2 + (1257 Ω − 1061 Ω) 2 ; Z = 358 Ω *32-30. What is the effective ac current delivered to the ac series circuit described in Problem 32-29? What is the peak value for this current? V 90 V i= = ; i = 300 mA Z 300 Ω imax = 2(300 mA); i = 424 mA 193
9. 9. Chapter 32. Alternating-Current Circuits Physics, 6th Edition *32-31. A series ac circuit consists of a 100-Ω resistor, a 0.2-H inductor, and a 3-µF capacitor connected to a 110-V, 60 Hz ac source. What is the inductive reactance, the capacitative reactance, and the impedance for the circuit? 1 XC = = 884 Ω; X L = 2π (60 Hz)(0.2 H) = 75.4 Ω 2π (60 Hz)(3 x 10-6 F) Z = R 2 + ( X L − X C ) 2 = (100 Ω) 2 + (75.4 Ω − 884 Ω) 2 ; Z = 815 Ω *32-32. What are the phase angle and the power factor for the circuit described in Problem 32-31? X L − X C 75.4 Ω − 884 Ω tan φ = = ; φ = -83.00 R 100 Ω Power Factor = cos φ = cos (-83.00); Power Factor = 0.123 or 12.3% *32-33. Assume that all the elements of Problem 32-28 above are connected in series with the given source. What is the effective current delivered to the circuit? From Problem 32-28: R = 20 Ω, XC = 1326, and XL = 264 Ω, V = 120 V, f = 60 Hz Z = R 2 + ( X L − X C ) 2 = (20 Ω) 2 + (264 Ω − 1326 Ω) 2 ; Z = 1062 Ω V 120 V i= = ; i = 113 mA Z 1062 Ω *32-34. A series ac circuit contains a 12-mH inductor, an 8-µF capacitor, and a 40-Ω resistor connected to a 110-V, 200 Hz ac line. What is the effective ac current in the circuit? 1 XC = = 99.5 Ω; X L = 2π (200 Hz)(0.012 H) = 15.1 Ω 2π (200 Hz)(8 x 10-6 F) Z = R 2 + ( X L − X C ) 2 = (40 Ω) 2 + (15.1 Ω − 99.5 Ω) 2 ; Z = 93.4 Ω 194
10. 10. Chapter 32. Alternating-Current Circuits Physics, 6th Edition X L − X C 15.1 Ω − 99.5 Ω *32-34. (Cont.) tan φ = = ; φ = -64.60 R 40 Ω Power Factor = cos φ = cos (-64.60); Power Factor = 0.428 or 42.8% *32-35. When a 6-Ω resistor and a pure inductor are connected to a 100-V, 60 Hz ac line, the effective current is 10 A. What is the inductance ? What is the power loss through the resistor and what is the power loss through the inductor? 110 V Z= = 11.0 Ω; X L = Z 2 − R 2 = (11.0) 2 − (6 Ω) 2 ; 10 A XL (9.22 Ω) XL = 9.22 Ω; L= = ; L = 24.5 mH 2π f 2π (60 Hz) P = I2R = (10 A)2 (6.00 Ω); P = 600 W; No Power lost in inductor: 0 *32-36. A capacitor is in series with a resistance of 35 Ω and connected to a 220-V, ac line. The reactance of the capacitor is 45 Ω. What is the effective ac current? What is the phase angle? What is the power factor? Z = R 2 + (0 − X C ) 2 = (35 Ω) 2 + (45 Ω) 2 = 57.0 Ω; 220 V i= = 3.89 A; i = 3.89 A 57.0 Ω X L − X C −45 Ω tan φ = = ; φ = -52.10; R 35 Ω Power Factor = cos 52.10 = 61.4% 195
11. 11. Chapter 32. Alternating-Current Circuits Physics, 6th Edition *32-37. A coil having an inductance of 0.15 H and a resistance of 12 Ω is connected to a 110-V, 25 Hz line. What is the effective current in the circuit? What is the power factor? What power is lost in the circuit? ( XC = 0 ) X L = 2π fL = 2π (25 Hz)(0.15 H) = 23.56 Ω; Z = R2 + X L 2 110 V Z = (12) 2 + (23.56) 2 = 26.44 Ω; i= ; i = 4.16 A 26.44 Ω X L − X C 23.56 Ω tan φ = = ; φ = 630; Power Factor = cos 630 = 45.4% R 12.0 Ω Power is lost only through resistance: P = I2R = (4.16 A)2(12 Ω) = 208 W *32-38. What is the resonant frequency for the circuit described by Problem 32-29? 1 1 fr = = ; fr = 205 Hz 2π LC 2π (0.2 H)(3 x 10-6 F) *32-39. What is the resonant frequency for the circuit described by Problem 32-34? 1 1 fr = = ; fr = 514 Hz 2π LC 2π (0.012 H)(8 x 10-6 F) Challenge Problems 32-40. A 100-V battery is connected to a dc series circuit with a 60-mH inductor and a 50-Ω resistor. Calculate the maximum current in the circuit and the time constant. VB V 100 V When t = ∞, i = (1 − e − ( R / L ) t ) = B ( 1 − 0 ) ; i = ; imax = 2.00 A R R 50 Ω L 0.060 H τ= = ; τ = 1.20 ms R 50 Ω 196
12. 12. Chapter 32. Alternating-Current Circuits Physics, 6th Edition 32-41. If the circuit of Problem 32-40 reaches a steady state condition and is then broken, what will be the instantaneous current after one time constant? (Decay of an inductor.) VB − ( R / L ) t 100 V  1  i= e =  ; i = 736 mA R 50 Ω  e  32-42. A resonant circuit has an inductance of 400 µH and a capacitance of 100 pF. What is the resonant frequency? 1 1 fr = = ; fr = 796 kHz 2π LC 2π (4 x 10-4 H)(1 x 10-10 F) 32-43. An LR dc circuit has a time constant of 2 ms. What is the inductance if the resistance is 2 kΩ? Find the instantaneous current 2 ms after the circuit is connected to a 12-V battery. L τ= ; L = τ R = (0.002 s)(2000 Ω); L = 4.00 H R VB 12 V 1 i= (1 − e − ( R / L ) t ) = (1 − ); i = 3.79 mA R 2000 Ω e 32-44. A series dc circuit consists of a 12-V battery, a 20-Ω resistor, and an unknown capacitor. The time constant is 40 ms. What is the capacitance? What is the maximum charge on the capacitor? τ 0.040 s τ = RC ; C = = ; C = 2.00 mF; R 20 Ω Q = CV = (2 mF)(12 V) = 24 mC 197
13. 13. Chapter 32. Alternating-Current Circuits Physics, 6th Edition *32-45. A 2-H inductor having a resistance of 120 Ω is connected to a 30-V battery. How much time is required for the current to reach 63 percent of its maximum value? What is the initial rate of current increase in amperes per second? What is the final current? The current reaches 63% of its maximum value in a time equal to one time constant: L 2H L R τ= = ; τ = 16.7 ms R 120 Ω ∆i For Loop: ΣE = ΣIR or VB − L = iR ∆t 0 ∆i ∆i VB Solving for ∆i/∆t with initial i = 0, we have: VB − iR = L ; = ; ∆t ∆t L ∆i VB 30 V ∆i = = ; = 15 A/s ∆t L 2H ∆t VB 30 V The final current is the maximum: imax = = ; imax = 250 mA R 120 Ω *32-46. Consider the circuit shown in Fig. 38-13. What is the impedance? What is the effective current? What is the power loss in the circuit? L R C L = 4 mH; C = 10 µF; R = 20 Ω; XL = 2πfL = 2π(600 Hz)(0.004 H) = 15.1 Ω Source: 50 V, 600 Hz 1 XC = = 26.5 Ω; Z = (20 Ω) 2 + (15.1 Ω − 26.5 Ω) 2 ; 2π (600 Hz)(10 x 10-6 F) V 50 V Impedance: Z = 23.0 Ω i= = ; Effective current: i = 2.17 A Z 23 Ω Power loss is entirely in resistance: P = i2R = (2.17 A)2(20 Ω); P = 94.2 W 198
14. 14. Chapter 32. Alternating-Current Circuits Physics, 6th Edition *32-47. A tuner circuit contains a 4-mH inductor and a variable capacitor. What must be the capacitance if the circuit is to resonate at a frequency of 800 Hz? 1 1 1 fr = ; f r2 = ; C= 2π LC 4π 2 LC 4π Lf r2 2 1 C= ; C = 9.89 µF 4π (0.004 H)(800 Hz) 2 2 *32-48. An 8-µF capacitor is in series with a 40-Ω resistor and connected to a 117-V, 60-Hz ac source. What is the impedance? What is the power factor? How much power is lost in the circuit? ( Note: XL = 0, so that all reactance is due to the capacitor. ) 1 XC = = 332 Ω 2π (60 Hz)(8 x 10-6 F) R C Z = (40 Ω) 2 + (0 − 332) 2 ; Z = 334 Ω Source: 117 V, 60 Hz V 117 V i= = ; i = 350 mA Z 334 Ω X L − X C -332 Ω tan φ = = ; φ =-83.10; Power Factor = cos 83.10 = 12.0% R 40 Ω Power is lost only through resistance: P = I2R = (0.350 A)2(40 Ω) = 4.90 W *32-49. Someone wants to construct a circuit whose resonant frequency is 950 kHz. If a coil in the circuit has an inductance of 3 mH, what capacitance should be added to the circuit? 1 1 1 fr = ; f r2 = ; C= 2π LC 4π 2 LC 4π Lf r2 2 1 C= ; C = 9.36 pF 4π (0.003 H)(950,000 Hz) 2 2 199
15. 15. Chapter 32. Alternating-Current Circuits Physics, 6th Edition *32-50. A 50-µF capacitor and a 70-Ω resistor are connected in series across a 120-V, 60 Hz ac line. Determine the current in the circuit, the phase angle, and the total power loss? 1 XC = = 53.05 Ω 2π (60 Hz)(50 x 10-6 F) R C Z = (70 Ω) 2 + (0 − 53.05) 2 ; Z = 87.8 Ω Source: 120 V, 60 Hz V 120 V i= = ; i = 1.37 A Z 87.8 Ω X L − X C -53.05 Ω tan φ = = ; φ =-37.20; Power Factor = cos 37.20 = 79.8% R 70 Ω Power is lost only through resistance: P = I2R = (1.37 A)2(70 Ω) = 131 W *32-51. Refer to Problem 32-50. What is the voltage across the resistor? What is the voltage across the capacitor? What inductance should be added to the circuit to reach resonance? V = iR = (1.37 A)(70 Ω) = 95.6 V VC = iXC = (1.37 A)(53.05 Ω) = 72.5 V 1 1 1 fr = ; f r2 = ; L= 2π LC 4π 2 LC 4π Cf r2 2 1 L= ; L = 141 mH 4π (50 x 10-6 F)(60 Hz) 2 2 200
16. 16. Chapter 32. Alternating-Current Circuits Physics, 6th Edition Critical Thinking Problems *32-52 The resistance of an 8-H inductor is 200 Ω. If this inductor is suddenly connected across a potential difference of 50 V, what is the initial rate of increase of the current in A/s (see Eq. 32-15)? What is the final steady current? At what rate is the current increasing after one time constant. At what time after connecting the source of emf, does the current equal one-half of its final value? (Initially, the current is zero.) L 8H L τ= = ; τ = 40.0 ms R R 200 Ω ∆i For Loop: ΣE = ΣIR or VB − L = iR ∆t 0 ∆i ∆i VB Solving for ∆i/∆t with initial i = 0, we have: VB − iR = L ; = ; ∆t ∆t L ∆i VB 50 V ∆i = = ; = 6.25 A/s ∆t L 8H ∆t VB 50 V The final current is the maximum: imax = = ; imax = 250 mA R 200 Ω To find the rate of increase after one time constant, we need to know i at t = τ:  1 i = imax (1 − e −t / τ ) = (250 mA) 1-  ; i = 158 mA  e ∆i ∆i ∆i VB − iR VB − L = iR; L = VB − iR; = ∆t ∆t ∆t L ∆i VB − iR 50 V - (0.158 A)(200 Ω) ∆i = = = 2.30 A/s ∆t L 8H ∆t Rt i Let x = , then : = (1 − e − x ) = 0.50; e − x = 0.5; e + x = 2.0 L imax Rt 0.693(8 H) x = 0.693 = ; t= ; t = 27.7 ms L 200 Ω 201
17. 17. Chapter 32. Alternating-Current Circuits Physics, 6th Edition *32-53. In Chapter 29, we found that the magnetic flux density within a solenoid was given by: Φ µ 0 NI (1) B = = A l Here we have used l for the length of the solenoid in order to avoid confusion with the symbol L used for inductance. We also know that induced emf is found in two ways: ∆Φ ∆i (2) E = − N and (3) E = − L ∆t ∆t Prove the following relationships for a solenoid of length l having N turns of area A: NΦ µN 2 A L= and L= I l Combining Eqs (2) and (3) above, we obtain: N ∆φ = L ∆i Nφ Now, at maximum current I: Nφ = LI or L= I µ0 NIA Nφ N  µ0 NIA  From (1): φ = which gives : L = =   l I I  l  µ0 N 2 A And finally: L= Note: L is not dependent on current or voltage. l *32-54. An inductor consists of a coil 30 cm long with 300 turns of area 0.004 m2. Find the inductance (see previous problem). A battery is connected and the current builds from zero to 2.00 A in 0.001 s. What is the average induced emf in the coil? µ0 N 2 A (4π x 10-7 T ⋅ m/A)(300) 2 (0.004 m 2 ) L= = ; L = 1.51 mH l 0.300 m ∆i 2A-0 E = −L = −(1.51 x 10-3 H)  ; E = -3.02 V ∆t  0.001 s  202
18. 18. Chapter 32. Alternating-Current Circuits Physics, 6th Edition *32-55. An inductor, resistor, and capacitor are connected in series with a 60-Hz ac line. A voltmeter connected to each element in the circuit gives the following readings: VR = 60 V, VL = 100 V, and VC = 160 V. What is the total voltage drop in the circuit? What is the phase angle? VT = VR2 + (VL − VC ) 2 = (60 V) 2 + (100 V - 160 V) 2 ; VT = 84.6 V VL − VC (100 V - 160 V) tan φ = = ; φ = -45.00 VR 60 V *32-56. The antenna circuit in a radio receiver consists of a variable capacitor and a 9-mH coil. The resistance of the circuit is 40 Ω. A 980-kHz radio waves produces a potential difference of 0.2 mV across the circuit. Determine the capacitance required for resonance. . What is the current at resonance? 1 1 1 fr = ; f r2 = ; C= 2π LC 4π 2 LC 4π Lf r2 2 1 C= ; C = 2.93 µF 4π (0.009 H)(980,000 Hz) 2 2 0.2 mV At fr , XC = XL and: Z = R = 40 Ω; i = ; i = 5.00 mA 40 Ω 203
19. 19. Chapter 32. Alternating-Current Circuits Physics, 6th Edition *32-57. A series RLC circuit has elements as follows: L = 0.6 H, C = 3.5 µF, and R = 250 Ω. The circuit is driven by an generator that produces a maximum emf of 150 V at 60 Hz. What is the effective ac current? What is the phase angle? What is the average power loss in the circuit? ( First find Veff ) L R C Vmax 150 V Veff = = ; Veff = 106 V 2 2 Source: 150 V, 60 Hz L = 0.600 H; C = 3.5 µF; R = 250 Ω; XL = 2πfL = 2π(60 Hz)(0.6 H) = 226 Ω 1 XC = = 758 Ω; Z = (250 Ω) 2 + (226 Ω − 758 Ω) 2 ; 2π (60 Hz)(3.5 x 10-6 F) V 106 V Impedance: Z = 588 Ω i= = ; Effective current: i = 181 mA Z 588 Ω X L − X C 226 Ω − 768 Ω tan φ = = ; φ = 42.70 R 250 Ω Power loss is entirely in resistance: P = i2R = (0.181 A)2(250 Ω); P = 8.19 W 204