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31 ALTERNATING CURRENT 31.1. IDENTIFY: i I cost and I rms I/ 2. SET UP: The specified value is the root-mean-square current; I rms 0.34 A. EXECUTE: (a) I rms 0.34 A 31.2. (b) I 2I rms 2(0.34 A) 0.48 A. (c) Since the current is positive half of the time and negative half of the time, its average value is zero. 2 (0.34 A)2 0.12 A 2. (d) Since I rms is the square root of the average of i 2, the average square of the current is I rms EVALUATE: The current amplitude is larger than its rms value. IDENTIFY and SET UP: Apply Eqs.(31.3) and (31.4) EXECUTE: (a) I 2I rms 2(2.10 A) 2.97 A. (b) I rav 31.3. 31.4. 2 I 2 (2.97 A) 1.89 A. EVALUATE: (c) The root-mean-square voltage is always greater than the rectified average, because squaring the current before averaging, and then taking the square root to get the root-mean-square value will always give a larger value than just averaging. IDENTIFY and SET UP: Apply Eq.(31.5). V 45.0 V EXECUTE: (a) Vrms 31.8 V. 2 2 (b) Since the voltage is sinusoidal, the average is zero. EVALUATE: The voltage amplitude is larger than Vrms . IDENTIFY: V IX C with X C 1 C . is the angular frequency, in rad/s. I EXECUTE: (a) V IX C so I V C (60.0 V)(100 rad s)(2.20 106 F) 0.0132 A. C (b) I V C (60.0 V)(1000 rad s)(2.20 106 F) 0.132 A. SET UP: (c) I V C (60.0 V)(10,000 rad s)(2.20 106 F) 1.32 A. (d) The plot of log I versus log is given in Figure 31.4. EVALUATE: I VC so log I log(VC) log. A graph of log I versus log should be a straight line with slope +1, and that is what Figure 31.4 shows. Figure 31.4 31-1 31-2 31.5. Chapter 31 IDENTIFY: V IX L with X L L . SET UP: is the angular frequency, in rad/s. V 60.0 V 0.120 A. EXECUTE: (a) V IX L I L and I L (100 rad s)(5.00 H) (b) I V 60.0 V 0.0120 A . L (1000 rad s)(5.00 H) (c) I V 60.0 V 0.00120 A . L (10,000 rad s)(5.00 H) (d) The plot of log I versus log is given in Figure 31.5. EVALUATE: I V L so log I log(V / L) log . A graph of log I versus log should be a straight line with slope 1, and that is what Figure 31.5 shows. Figure 31.5 31.6. IDENTIFY: The reactance of capacitors and inductors depends on the angular frequency at which they are operated, as well as their capacitance or inductance. SET UP: The reactances are X C 1/ C and X L L . EXECUTE: (a) Equating the reactances gives L (b) Using the numerical values we get 31.7. 1 C 1 LC 1 1 = 7560 rad/s LC (5.00 mH)(3.50 µF) XC = XL = L = (7560 rad/s)(5.00 mH) = 37.8 Ω EVALUATE: At other angular frequencies, the two reactances could be very different. IDENTIFY and SET UP: For a resistor vR iR. For an inductor, vL V cos(t 90°). For a capacitor, vC V cos(t 90°). EXECUTE: The graphs are sketched in Figures 31.7a-c. The phasor diagrams are given in Figure 31.7d. EVALUATE: For a resistor only in the circuit, the current and voltage in phase. For an inductor only, the voltage leads the current by 90°. For a capacitor only, the voltage lags the current by 90°. Figure 31.7a and b NAlternating Current 31-3 Figure 31.7c Figure 31.7d 31.8. IDENTIFY: The reactance of an inductor is X L L 2 fL . The reactance of a capacitor is X C 1 C 1 . 2 fC SET UP: The frequency f is in Hz. EXECUTE: (a) At 60.0 Hz, X L 2 (60.0 Hz)(0.450 H) 170 . X L is proportional to f so at 600 Hz, X L 1700 . (b) At 60.0 Hz, X C 1 1.06 103 . X C is proportional to 1/ f , so at 600 Hz, X C 106 . 2 (60.0 Hz)(2.50 106 F) (c) X L X C says 2 fL 31.9. EVALUATE: X L increases when f increases. X C increases when f increases. IDENTIFY and SET UP: Use Eqs.(31.12) and (31.18). EXECUTE: (a) X L L 2 fL 2 (80.0 Hz)(3.00 H) 1510 (b) X L 2 fL gives L (c) X C (d) X C 1 C IDENTIFY: SET UP: 1 1 497 2 fC 2 (80.0 Hz)(4.00 106 F) X L increases when L increases; X C decreases when C increases. VL I L is the angular frequency, in rad/s. f is the frequency in Hz. 2 VL (12.0 V) 1.63 106 Hz. 2 IL 2 (2.60 103 A)(4.50 104 H) EVALUATE: When f is increased, I decreases. IDENTIFY and SET UP: Apply Eqs.(31.18) and (31.19). V 170 V EXECUTE: V IX C so X C 200 I 0.850 A 1 1 1 XC gives C 1.33 105 F 13.3 F C 2 f X C 2 (60.0 Hz)(200 ) EVALUATE: The reactance relates the voltage amplitude to the current amplitude and is similar to Ohm’s law. EXECUTE: 31.11. XL 120 0.239 H 2 f 2 (80.0 Hz) 1 1 1 gives C 1.66 105 F 2 fC 2 f X C 2 (80.0 Hz)(120 ) EVALUATE: 31.10. 1 1 1 and f 150 Hz. 2 fC 2 LC 2 (0.450 H)(2.50 106 F) VL I L so f 31-4 Chapter 31 31.12. IDENTIFY: Compare vC that is given in the problem to the general form vC I sin t and determine . C 1 . vR iR and i I cos. C 1 1 1736 EXECUTE: (a) X C C (120 rad s)(4.80 106 F) V 7.60 V 4.378 103 A and i I cos t (4.378 103 A)cos[(120 rad/s)t ]. Then (b) I C X C 1736 SET UP: 31.13. 31.14. 31.15. XC vR iR (4.38 103 A)(250 )cos((120 rad s)t ) (1.10 V)cos((120 rad s)t ). EVALUATE: The voltage across the resistor has a different phase than the voltage across the capacitor. IDENTIFY and SET UP: The voltage and current for a resistor are related by vR iR. Deduce the frequency of the voltage and use this in Eq.(31.12) to calculate the inductive reactance. Eq.(31.10) gives the voltage across the inductor. EXECUTE: (a) vR (3.80 V)cos[(720 rad/s)t ] v 3.80 V vR iR, so i R cos[(720 rad/s)t ] (0.0253 A)cos[(720 rad/s)t ] R 150 (b) X L L 720 rad/s, L 0.250 H, so X L L (720 rad/s)(0.250 H) 180 (c) If i I cost then vL VL cos(t 90) (from Eq.31.10). VL I L IX L (0.02533 A)(180 ) 4.56 V vL (4.56 V)cos[(720 rad/s)t 90] But cos(a 90) sin a (Appendix B), so vL (4.56 V)sin[(720 rad/s)t ]. EVALUATE: The current is the same in the resistor and inductor and the voltages are 90 out of phase, with the voltage across the inductor leading. IDENTIFY: Calculate the reactance of the inductor and of the capacitor. Calculate the impedance and use that result to calculate the current amplitude. X V SET UP: With no capacitor, Z R 2 X L2 and tan L . X L L. I . VL IX L and VR IR. For an R Z inductor, the voltage leads the current. EXECUTE: (a) X L L (250 rad/s)(0.400 H) 100 . Z (200 )2 (100 ) 2 224 . V 30.0 V (b) I 0.134 A Z 224 (c) VR IR (0.134 A)(200 ) 26.8 V. VL IX L (0.134 A)(100 ) 13.4 V. X 100 (d) tan L and 26.6°. Since is positive, the source voltage leads the current. R 200 (e) The phasor diagram is sketched in Figure 31.14. EVALUATE: Note that VR VL is greater than V. The loop rule is satisfied at each instance of time but the voltages across R and L reach their maxima at different times. Figure 31.14 IDENTIFY: vR (t ) is given by Eq.(31.8). vL (t ) is given by Eq.(31.10). SET UP: From Exercise 31.14, V 30.0 V , VR 26.8 V , VL 13.4 V and 26.6° . EXECUTE: (a) The graph is given in Figure 31.15. (b) The different voltages are v (30.0 V)cos(250t 26.6), vR (26.8 V)cos(250t ), vL (13.4 V)cos(250t 90). At t 20 ms: v 20.5 V, vR 7.60 V, vL 12.85 V. Note that vR vL v. NAlternating Current 31-5 (c) At t 40 ms: v 15.2 V, vR 22.49 V, vL 7.29 V. Note that vR vL v. EVALUATE: It is important to be careful with radians versus degrees in above expressions! 31.16. Figure 31.15 IDENTIFY: Calculate the reactance of the inductor and of the capacitor. Calculate the impedance and use that result to calculate the current amplitude. X XC 1 SET UP: With no resistor, Z ( X L X C ) 2 X L X C . tan L . XC . X L L. For an zero C inductor, the voltage leads the current. For a capacitor, the voltage lags the current. 1 1 667 . EXECUTE: (a) X L L (250 rad/s)(0.400 H) 100 . X C C (250 rad/s)(6.00 106 F) Z X L X C 100 667 567 . V 30.0 V 0.0529 A Z 567 (c) VC IX C (0.0529 A)(667 ) 35.3 V. VL IX L (0.0529 A)(100 ) 5.29 V. X X C 100 667 (d) tan L and 90°. The phase angle is negative and the source voltage lags zero zero the current. (e) The phasor diagram is sketched in Figure 31.16. EVALUATE: When X C X L the phase angle is negative and the source voltage lags the current. (b) I Figure 31.16 31.17. IDENTIFY and SET UP: Calculate the impendance of the circuit and use Eq.(31.22) to find the current amplitude. The voltage amplitudes across each circuit element are given by Eqs.(31.7), (31.13), and (31.19). The phase angle is calculated using Eq.(31.24). The circuit is shown in Figure 31.17a. No inductor means X L 0 R 200 , C 6.00 106 F, V 30.0 V, 250 rad/s Figure 31.17a 31-6 Chapter 31 EXECUTE: (a) X C 1 1 666.7 C (250 rad/s)(6.00 106 F) Z R 2 ( X L X C ) 2 (200 ) 2 (666.7 ) 2 696 V 30.0 V 0.0431 A 43.1 mA Z 696 (c) Voltage amplitude across the resistor: VR IR (0.0431 A)(200 ) 8.62 V Voltage amplitude across the capacitor: VC IX C (0.0431 A)(666.7 ) 28.7 V X X C 0 666.7 (d) tan L 3.333 so 73.3 R 200 The phase angle is negative, so the source voltage lags behind the current. (e) The phasor diagram is sketched qualitatively in Figure 31.17b. (b) I 31.18. 31.19. Figure 31.17b EVALUATE: The voltage across the resistor is in phase with the current and the capacitor voltage lags the current by 90. The presence of the capacitor causes the source voltage to lag behind the current. Note that VR VC V . The instantaneous voltages in the circuit obey the loop rule at all times but because of the phase differences the voltage amplitudes do not. IDENTIFY: vR (t ) is given by Eq.(31.8). vC (t ) is given by Eq.(31.16). SET UP: From Exercise 31.17, V 30.0 V, VR 8.62 V, VC 28.7 V and 73.3°. EXECUTE: (a) The graph is given in Figure 31.18. (b) The different voltage are: v (30.0 V)cos(250t 73.3), vR (8.62 V)cos(250t ), vC (28.7 V)cos(250t 90). At t 20 ms : v 25.1 V, vR 2.45 V, vC 27.5 V. Note that vR vC v. (c) At t 40 ms: v 22.9 V, vR 7.23 V, vC 15.6 V. Note that vR vC v. EVALUATE: It is important to be careful with radians vs. degrees! Figure 31.18 IDENTIFY: Apply the equations in Section 31.3. SET UP: 250 rad/s, R 200 , L 0.400 H, C 6.00 F and V 30.0 V. EXECUTE: (a) Z R2 ( L 1/ C )2 . Z (200 )2 ((250 rad/s)(0.0400 H) 1/((250 rad/s)(6.00 106 F)))2 601 V 30 V 0.0499 A. (b) I Z 601 100 667 L 1/ C (c) arctan 70.6, and the voltage lags the current. arctan R 200 NAlternating Current 31-7 (d) VR IR (0.0499 A)(200 ) 9.98 V; VL I L (0.0499 A)(250 rad s)(0.400 H) 4.99 V; VC I (0.0499 A) 33.3 V. C (250 rad/s)(6.00 106 F) EVALUATE: (e) At any instant, v vR vC vL . But vC and v L are 180° out of phase, so vC can be larger than v at a value of t, if vL vR is negative at that t. 31.20. IDENTIFY: vR (t ) is given by Eq.(31.8). vC (t ) is given by Eq.(31.16). vL (t ) is given by Eq.(31.10). SET UP: From Exercise 31.19, V 30.0 V, VL 4.99 V, VR 9.98 V, VC 33.3 V and 70.6°. EXECUTE: (a) The graph is sketched in Figure 31.20. The different voltages plotted in the graph are: v (30 V)cos(250t 70.6), vR (9.98 V)cos(250t ), vL (4.99 V)cos(250t 90) and vC (33.3 V)cos(250t 90). (b) At t 20 ms: v 24.3 V, vR 2.83 V, vL 4.79 V, vC 31.9 V. (c) At t 40 ms: v 23.8 V, vR 8.37 V, vL 2.71 V, vC 18.1 V. EVALUATE: In both parts (b) and (c), note that the source voltage equals the sum of the other voltages at the given instant. Be careful with degrees versus radians! Figure 31.20 31.21. IDENTIFY and SET UP: The current is largest at the resonance frequency. At resonance, X L X C and Z R. For part (b), calculate Z and use I V / Z . 1 EXECUTE: (a) f 0 113 Hz. I V / R 15.0 mA. 2 LC (b) X C 1/ C 500 . X L L 160 . Z R 2 ( X L X C ) 2 (200 ) 2 (160 500 ) 2 394.5 . I V / Z 7.61mA. X C X L so the source voltage lags the current. 31.22. EVALUATE: 0 2 f0 710 rad/s. 400 rad/s and is less than 0 . When 0 , X C X L . Note that I in part (b) is less than I in part (a). IDENTIFY: The impedance and individual reactances depend on the angular frequency at which the circuit is driven. 2 1 SET UP: The impedance is Z R 2 L , the current amplitude is I = V/Z, and the instantaneous C values of the potential and current are v = V cos(t + ), where tan = (XL – XC)/R, and i = I cos t. 1 1 1 EXECUTE: (a) Z is a minimum when L , which gives = 3162 rad/s = C LC (8.00 mH)(12.5 µF) 3160 rad/s and Z = R = 175 Ω. (b) I = V/Z = (25.0 V)/(175 Ω) = 0.143 A (c) i = I cos t = I/2, so cost = 1 2 , which gives t = 60° = π/3 rad. v = V cos(t + ), where tan = (XL – XC)/R = 0/R =0. So, v = (25.0 V) cost = (25.0 V)(1/2) = 12.5 V. vR = Ri = (175 Ω)(1/2)(0.143 A) = 12.5 V. 0.143 A cos(60 90) = +3.13 V. vC = VC cos(t – 90°) = IXC cos(t – 90°) = (3162 rad/s)(12.5 µF) 31-8 Chapter 31 31.23. vL = VL cos(t + 90°) = IXL cos(t + 90°) = (0.143 A)(3162 rad/s)(8.00 mH) cos(60° + 90°). vL = –3.13 V. (d) vR + vL + vC = 12.5 V + (–3.13 V) + 3.13 V = 12.5 V = vsource EVALUATE: The instantaneous potential differences across all the circuit elements always add up to the value of the source voltage at that instant. In this case (resonance), the potentials across the inductor and capacitor have the same magnitude but are 180° out of phase, so they add to zero, leaving all the potential difference across the resistor. IDENTIFY and SET UP: Use the equation that preceeds Eq.(31.20): V 2 VR2 (VL VC )2 31.24. EXECUTE: V (30.0 V)2 (50.0 V 90.0 V) 2 50.0 V EVALUATE: The equation follows directly from the phasor diagrams of Fig.31.13 (b or c). Note that the voltage amplitudes do not simply add to give 170.0 V for the source voltage. 1 IDENTIFY and SET UP: X L L and X C . C 1 L 1 1 EXECUTE: (a) If 0 , then X L and X 0. C LC LC C LC (b) When 0 , X 0 (c) When 0 , X 0 (d) The graph of X versus is given in Figure 31.24. EVALUATE: Z R 2 X 2 and tan X / R. Figure 31.24 31.25. 2 R. IDENTIFY: For a pure resistance, Pav Vrms I rms I rms SET UP: 20.0 W is the average power Pav . EXECUTE: (a) The average power is one-half the maximum power, so the maximum instantaneous power is 40.0 W. P 20.0 W 0.167 A (b) I rms av Vrms 120 V P 20.0 W 720 (c) R 2av I rms (0.167 A) 2 31.26. VR2,rms 2 Vrms (120 V) 2 20.0 W. R R 750 IDENTIFY: The average power supplied by the source is P Vrms I rms cos . The power consumed in the resistance EVALUATE: We can also calculate the average power as Pav 2 R. is P I rms SET UP: 2 f 2 (1.25 103 Hz) 7.854 103 rad/s. X L L 157 . X C 1 C 909 . EXECUTE: (a) First, let us find the phase angle between the voltage and the current: X X C 157 909 tan L and 65.04. The impedance of the circuit is R 350 Z R 2 ( X L X C ) 2 (350 ) 2 (752 ) 2 830 . The average power provided by the generator is then P Vrms I rms cos( ) 2 Vrms (120 V)2 cos( ) cos(65.04) 7.32 W Z 830 NAlternating Current 31-9 2 31.27. 120 V 2 (b) The average power dissipated by the resistor is PR I rms R (350 ) 7.32 W. 830 EVALUATE: Conservation of energy requires that the answers to parts (a) and (b) are equal. IDENTIFY: The power factor is cos , where is the phase angle in Fig.31.13. The average power is given by Eq.(31.31). Use the result of part (a) to rewrite this expression. (a) SET UP: The phasor diagram is sketched in Figure 31.27. EXECUTE: From the diagram V IR R cos R , V IZ Z as was to be shown. 31.28. 31.29. Figure 31.27 V R V 2 (b) Pav Vrms I rms cos Vrms I rms rms I rms R. But rms I rms , so Pav I rms R. Z Z Z EVALUATE: In an L-R-C circuit, electrical energy is stored and released in the inductor and capacitor but none is dissipated in either of these circuit elements. The power delivered by the source equals the power dissipated in the resistor. V R IDENTIFY and SET UP: Pav Vrms I rms cos . I rms rms . cos . Z Z 80.0 V 75.0 EXECUTE: I rms 0.762 A. cos 0.714. Pav (80.0 V)(0.762 A)(0.714) 43.5 W. 105 105 EVALUATE: Since the average power consumed by the inductor and by the capacitor is zero, we can also 2 R (0.762 A) 2 (75.0 ) 43.5 W. calculate the average power as Pav I rms IDENTIFY and SET UP: Use the equations of Section 31.3 to calculate , Z and Vrms . The average power 2 R delivered by the source is given by Eq.(31.31) and the average power dissipated in the resistor is I rms EXECUTE: (a) X L L 2 f L 2 (400 Hz)(0.120 H) 301.6 1 1 1 XC 54.51 C 2 fC 2 (400 Hz)(7.3 106 Hz) X X C 301.6 54.41 tan L , so 45.8. The power factor is cos 0.697. R 240 (b) Z R 2 ( X L X C ) 2 (240 ) 2 (301.6 54.51 ) 2 344 (c) Vrms I rms Z (0.450 A)(344 ) 155 V (d) Pav I rmsVrms cos (0.450 A)(155 V)(0.697) 48.6 W 31.30. 2 R (0.450 A)2 (240 ) 48.6 W (e) Pav I rms EVALUATE: The average electrical power delivered by the source equals the average electrical power consumed in the resistor. (f ) All the energy stored in the capacitor during one cycle of the current is released back to the circuit in another part of the cycle. There is no net dissipation of energy in the capacitor. (g) The answer is the same as for the capacitor. Energy is repeatedly being stored and released in the inductor, but no net energy is dissipated there. IDENTIFY: The angular frequency and the capacitance can be used to calculate the reactance X C of the capacitor. The angular frequency and the inductance can be used to calculate the reactance X L of the inductor. Calculate the phase angle and then the power factor is cos. Calculate the impedance of the circuit and then the rms current in the circuit. The average power is Pav Vrms I rms cos . On the average no power is consumed in the capacitor or the inductor, it is all consumed in the resistor. V 45 V SET UP: The source has rms voltage Vrms 31.8 V. 2 2 31-10 Chapter 31 EXECUTE: X L L (360 rad/s)(15 103 H) 5.4 . X C 1 1 794 . C (360 rad/s)(3.5 106 F) X L X C 5.4 794 and 72.4°. The power factor is cos 0.302. R 250 V 31.8 V (b) Z R 2 ( X L X C ) 2 (250 ) 2 (5.4 794 ) 2 827 . I rms rms 0.0385 A. Z 827 Pav Vrms I rms cos (31.8 V)(0.0385 A)(0.302) 0.370 W. tan 31.31. 2 R (0.0385 A) 2 (250 ) 0.370 W. The average power (c) The average power delivered to the resistor is Pav I rms delivered to the capacitor and to the inductor is zero. EVALUATE: On average the power delivered to the circuit equals the power consumed in the resistor. The capacitor and inductor store electrical energy during part of the current oscillation but each return the energy to the circuit during another part of the current cycle. IDENTIFY and SET UP: At the resonance frequency, Z = R. Use that V = IZ, VR IR, VL IX L and VC IX C . Pav is given by Eq.(31.31). (a) EXECUTE: V IZ IR (0.500 A)(300 ) 150 V (b) VR IR 150 V X L L L(1/ LC ) L / C 2582 ; VL IX L 1290 V X C 1/(C) L / C 2582 ; VC IX C 1290 V (c) Pav 12 VI cos 12 I 2 R, since V IR and cos 1 at resonance. Pav 12 (0.500 A) 2 (300 ) 37.5 W EVALUATE: At resonance VL VC . Note that VL VC V . However, at any instant vL vC 0. 31.32. 31.33. IDENTIFY: The current is maximum at the resonance frequency, so choose C such that 50.0 rad/s is the resonance frequency. At the resonance frequency Z R. SET UP: VL I L V EXECUTE: (a) The amplitude of the current is given by I . Thus, the current will have a 2 R ( L 1 ) 2 C 1 1 1 . Therefore, C 2 maximum amplitude when L 44.4 F. C L (50.0 rad/s) 2 (9.00 H) (b) With the capacitance calculated above we find that Z R, and the amplitude of the current is 120 V I V 0.300 A. Thus, the amplitude of the voltage across the inductor is R 400 VL I ( L) (0.300 A)(50.0 rad/s)(9.00 H) 135 V. EVALUATE: Note that VL is greater than the source voltage amplitude. IDENTIFY and SET UP: At resonance X L X C , 0 and Z R. R 150 , L 0.750 H, C 0.0180 F, V 150 V EXECUTE: (a) At the resonance frequency X L X C and from tan power factor is cos 1.00. (b) Pav 12 VI cos (Eq.31.31) At the resonance frequency Z = R, so I 31.34. X L XC we have that 0 and the R V V Z R V 2 1 (150 V)2 V Pav 12 V cos 12 2 75.0 W R 150 R (c) EVALUATE: When C and f are changed but the circuit is kept on resonance, nothing changes in Pav V 2 /(2 R), so the average power is unchanged: Pav 75.0 W. The resonance frequency changes but since Z = R at resonance the current doesn’t change. 1 IDENTIFY: 0 . VC IX C . V IZ . LC SET UP: At resonance, Z R. NAlternating Current EXECUTE: (a) 0 31.35. 31-11 1 1 1.54 104 rad/s LC (0.350 H)(0.0120 106 F) V V 1 1 5.41 103 . (b) V IZ C Z C R. X C 4 C (1.54 10 rad/s)(0.0120 106 F) XC XC 550 V V (400 ) 40.7 V. 3 5.4110 EVALUATE: The voltage amplitude for the capacitor is more than a factor of 10 times greater than the voltage amplitude of the source. 1 1 IDENTIFY and SET UP: The resonance angular frequency is 0 and . X L L. X C C LC Z R 2 ( X L X C ) 2 . At the resonance frequency X L X C and Z R. EXECUTE: (a) Z R 115 1 (b) 0 1.33 104 rad/s . 2 0 2.66 104 rad/s. 3 6 (4.50 10 H)(1.26 10 F) X L L (2.66 104 rad/s)(4.50 103 H) 120 . X C 1 1 30 4 C (2.66 10 rad/s)(1.25 106 F) Z (115 )2 (120 30 ) 2 146 (c) 0 / 2 6.65 103 rad/s. X L 30 . X C 31.36. 1 C 120 . Z (115 )2 (30 120 )2 146 , the same value as in part (b). EVALUATE: For 20 , X L X C . For 0 / 2, X L X C . But ( X L X C ) 2 has the same value at these two frequencies, so Z is the same. IDENTIFY: At resonance Z R and X L X C . 1 . V IZ . VR IR, VL IX L and VC VL . LC 1 1 945 rad s. EXECUTE: (a) 0 LC 0.280 H 4.00 106 F SET UP: 0 (b) I = 1.20 A at resonance, so R Z V 120 V 70.6 I 1.70 A (c) At resonance, VR 120 V, VL VC I L 1.70 A 945 rad s 0.280 H 450 V. 31.37. EVALUATE: At resonance, VR V and VL VC 0. IDENTIFY and SET UP: Eq.(31.35) relates the primary and secondary voltages to the number of turns in each. I = 2 2 Vrms / R. V/R and the power consumed in the resistive load is I rms EXECUTE: (a) V2 N 2 N V 120 V so 1 1 10 V1 N1 N 2 V2 12.0 V V2 12.0 V 2.40 A R 5.00 (c) Pav I 22 R (2.40 A) 2 (5.00 ) 28.8 W (d) The power drawn from the line by the transformer is the 28.8 W that is delivered by the load. (b) I 2 Pav 2 V2 V 2 (120 V)2 so R 500 R Pav 28.8 W N And 1 (5.00 ) (10) 2 (5.00 ) 500 , as was to be shown. N2 EVALUATE: The resistance is “transformed”. A load of resistance R connected to the secondary draws the same power as a resistance ( N1 / N 2 ) 2 R connected directly to the supply line, without using the transformer. 31-12 Chapter 31 31.38. IDENTIFY: SET UP: Pav VI and Pav,1 Pav,2 . V1 120 V. V2 13,000 V. EXECUTE: (a) 31.39. N1 V1 . N 2 V2 N 2 V2 13,000 V 108 N1 V1 120 V (b) Pav V2 I 2 (13,000 V)(8.50 103 A) 110 W P 110 W 0.917 A (c) I1 av V1 120 V EVALUATE: Since the power supplied to the primary must equal the power delivered by the secondary, in a stepup transformer the current in the primary is greater than the current in the secondary. V N IDENTIFY: A transformer transforms voltages according to 2 2 . The effective resistance of a secondary V1 N1 R V2 . Resistance R is related to Pav and V by Pav circuit of resistance R is Reff . Conservation of energy 2 ( N 2 / N1 ) R requires Pav,1 Pav,2 so V1I1 V2 I 2 . SET UP: Let V1 240 V and V2 120 V, so P2,av 1600 W. These voltages are rms. 31.40. EXECUTE: (a) V1 240 V and we want V2 120 V, so use a step-down transformer with N2 / N1 12 . P 1600 W (b) Pav VI , so I av 6.67 A. V 240 V V 2 (120 V)2 (c) The resistance R of the blower is R 9.00 . The effective resistance of the blower is P 1600 W 9.00 Reff 36.0 . (1/ 2) 2 EVALUATE: I 2V2 (13.3 A)(120 V) 1600 W. Energy is provided to the primary at the same rate that it is consumed in the secondary. Step-down transformers step up resistance and the current in the primary is less than the current in the secondary. 1 IDENTIFY: Z R 2 ( X L X C ) 2 , with X L L and X C . C SET UP: The woofer has a R and L in series and the tweeter has a R and C in series. EXECUTE: (a) Z tweeter R 2 (1 C )2 (b) Zwoofer R2 L 2 (c) If Z tweeter Z woofer , then the current splits evenly through each branch. (d) At the crossover point, where currents are equal, R 2 1 C 2 R 2 L . 2 f 31.41. 31.42. 1 . 2 2 LC 1 and LC EVALUATE: The crossover frequency corresponds to the resonance frequency of a R-C-L circuit, since the crossover frequency is where X L X C . IDENTIFY and SET UP: Use Eq.(31.24) to relate L and R to . The voltage across the coil leads the current in it by 52.3, so 52.3. X XC X EXECUTE: tan L . But there is no capacitance in the circuit so X C 0. Thus tan L and X L R R XL 62.1 0.124 H. R tan (48.0 )tan52.3 62.1 . X L L 2 f L so L 2 f 2 (80.0 Hz) EVALUATE: 45 when ( X L X C ) R, which is the case here. IDENTIFY: SET UP: Z R 2 ( X L X C ) 2 . I rms Vrms V 30.0 V 21.2 V. 2 2 Vrms . Vrms I rms R. VC ,rms I rms X C . VL ,rms I rms X L . Z NAlternating Current 31-13 EXECUTE: (a) 200 rad/s , so X L L (200 rad/s)(0.400 H) 80.0 and XC 1 1 833 . Z (200 )2 (80.0 833 ) 2 779 . C (200 rad/s)(6.00 106 F) I rms Vrms 21.2 V 0.0272 A. V1 reads VR ,rms I rms R (0.0272 A)(200 ) 5.44 V. V2 reads Z 779 I rms X L (0.0272 A)(80.0 ) 2.18 V. V3 reads VC ,rms I rms X C (0.0272 A)(833 ) 22.7 V. V4 reads VL ,rms VL VC VL,rms VC ,rms 2.18 V 22.7 V 20.5 V. V5 reads Vrms 21.2 V. 2 1 833 (b) 1000 rad/s so X L L (5)(80.0 ) 400 and X C 167 . C 5 V 21.2 V Z (200 )2 (400 167 )2 307 . I rms rms 0.0691 A. V1 reads VR ,rms 13.8 V. V2 reads Z 307 VL ,rms 27.6 V. V3 reads VC ,rms 11.5 V. V4 reads VL,rms VC ,rms 27.6 V 11.5 V 16.1 V. V5 reads Vrms 21.2 V. 1 645 rad/s. 200 rad/s is less than the LC resonance frequency and X C X L . 1000 rad/s is greater than the resonance frequency and X L X C . IDENTIFY and SET UP: The rectified current equals the absolute value of the current i. Evaluate the integral as specified in the problem. EXECUTE: (a) From Fig.31.3b, the rectified current is zero at the same values of t for which the sinusoidal current is zero. At these t, cos t 0 and t / 2, 3 / 2, . The two smallest positive times are t1 / 2 , t2 3 / 2. EVALUATE: The resonance frequency for this circuit is 0 31.43. (b) A t2 t1 t 2 t2 I 1 idt I costdt I sin t (sin t2 sin t1 ) t1 t1 sin t1 sin[ ( / 2 )] sin( / 2) 1 sin t2 sin[ (3 / 2 )] sin(3 / 2) 1 2I I A (1 (1)) (c ) I rav (t2 t1 ) 2I / I rav 31.44. 2I 2I (b) Z R2 X L 2 Vrms Z 400 250 2 2 472 . cos V V2 R R and I rms rms . Pav rms , so Z Z Z Z Pav 800 W 472 668 V. R 400 EVALUATE: 31.45. 2I , which is Eq.(31.3). (t2 t1 ) (3 / 2 / 2 ) EVALUATE: We have shown that Eq.(31.3) is correct. The average rectified current is less than the current amplitude I, since the rectified current varies between 0 and I. The average of the current is zero, since it has both positive and negative values. IDENTIFY: X L L. Pav Vrms I rms cos SET UP: f 120 Hz; 2 f . X 250 EXECUTE: (a) X L L L L 0.332 2 120 Hz I rms Vrms 668 V 2 R (1.415 A) 2 (400 ) 800 W, which 1.415 A. We can calculate Pav as I rms Z 472 checks. (a) IDENTIFY and SET UP: Source voltage lags current so it must be that X C X L and we must add an inductor in series with the circuit. When X C X L the power factor has its maximum value of unity, so calculate the additional L needed to raise X L to equal X C . 31-14 Chapter 31 (b) EXECUTE: power factor cos equals 1 so 0 and X C X L . Calculate the present value of X C X L to see how much more X L is needed: R Z cos (60.0 )(0.720) 43.2 X L XC so X L X C R tan R cos 0.720 gives 43.95 ( is negative since the voltage lags the current) tan Then X L X C R tan (43.2 ) tan(43.95) 41.64 . Therefore need to add 41.64 of X L . XL 41.64 0.133 H, amount of inductance to add. 2 f 2 (50.0 Hz) EVALUATE: From the information given we can’t calculate the original value of L in the circuit, just how much to add. When this L is added the current in the circuit will increase. 2 R IDENTIFY: Use Vrms I rms Z to calculate Z and then find R. Pav I rms X L L 2 f L and L 31.46. SET UP: X C 50.0 EXECUTE: R Z Vrms 240 V 2 80.0 R 2 X C2 R 2 50.0 . Thus, I rms 3.00 A 80.0 50.0 2 2 62.4 . The average power supplied to this circuit is equal to the power dissipated 2 by the resistor, which is P I rms R 3.00 A 62.4 562 W. 2 X L X C 50.0 and 38.7°. R 62.4 Pav Vrms I rms cos (240 V)(3.00 A)cos(38.7°) 562 W, which checks. IDENTIFY: The voltage and current amplitudes are the maximum values of these quantities, not necessarily the instantaneous values. SET UP: The voltage amplitudes are VR = RI, VL = XLI, and VC = XCI, where I = V/Z and EVALUATE: 31.47. tan 2 1 Z R2 L . C EXECUTE: (a) = 2πf = 2π(1250 Hz) = 7854 rad/s. Carrying extra figures in the calculator gives XL = L = (7854 rad/s)(3.50 mH) = 27.5 Ω; XC = 1/C = 1/[(7854 rad/s)(10.0 µF)] = 12.7 Ω; Z R 2 ( X L X C )2 = (50.0 )2 (27.5 12.7 ) 2 = 52.1 Ω; I = V/Z = (60.0 V)/(52.1 Ω) = 1.15 A; VR = RI = (50.0 Ω)(1.15 A) = 57.5 V; VL = XLI = (27.5 Ω)(1.15 A) = 31.6 V; VC = XCI = (12.7 Ω)(1.15 A) = 14.7 V. The voltage amplitudes can add to more than 60.0 V because these voltages do not all occur at the same instant of time. At any instant, the instantaneous voltages all add to 60.0 V. (b) All of them will change because they all depend on . XL = L will double to 55.0 Ω, and XC = 1/C will 31.48. decrease by half to 6.35 Ω. Therefore Z (50.0 )2 (55.0 6.35 ) 2 = 69.8 Ω; I = V/Z = (60.0 V)/(69.8 Ω) = 0.860 A; VR = IR = (0.860 A)(50.0 Ω) = 43.0 V; VL = IXL = (0.860 A)(55.0 Ω) = 47.3 V; VC = IXC = (0.860 A)(6.35 Ω) = 5.47 V. EVALUATE: The new amplitudes in part (b) are not simple multiples of the values in part (a) because the impedance and reactances are not all the same simple multiple of the angular frequency. 1 IDENTIFY and SET UP: X C . X L L. C 1 XL L 1 1 1L and LC 2 . At angular frequency 2 , 2 22 LC (21 ) 2 2 4. EXECUTE: (a) 1C 1 X C 1/ 2C 1 X L XC. XL 1 1 32 LC 1 2 . X C X L . XC 3 1 9 When increases, X L increases and X C decreases. When decreases, X L decreases and X C 2 (b) At angular frequency 3 , EVALUATE: increases. (c) The resonance angular frequency 0 is the value of for which X C X L , so 0 1. NAlternating Current 31.49. 31-15 IDENTIFY and SET UP: Express Z and I in terms of , L, C and R. The voltages across the resistor and the inductor are 90 out of phase, so Vout VR2 VL2 . EXECUTE: The circuit is sketched in Figure 31.49. X L L, X C 1 C 2 1 Z R2 L C V Vs I s 2 Z 1 R2 L C Figure 31.49 Vout I R 2 X L2 I R 2 2 L2 Vs Vout Vs R 2 2 L2 1 R2 L C 2 R 2 2 L2 1 R2 L C 2 small 2 1 1 2 2 2 2 As gets small, R 2 L 2 2 ,R L R C C Therefore Vout R2 RC as becomes small. Vs (1/ 2C 2 ) large 2 1 2 2 2 2 2 2 2 2 2 2 As gets large, R2 L R L L , R L L C Therefore, 31.50. Vout 2 L2 1 as becomes large. Vs 2 L2 EVALUATE: Vout / Vs 0 as becomes small, so there is Vout only when the frequency of Vs is large. If the source voltage contains a number of frequency components, only the high frequency ones are passed by this filter. IDENTIFY: V VC IX C . I V / Z . SET UP: EXECUTE: X L L, X C Vout VC 1 C . I V 1 out . 2 C Vs C R L 1 C 2 If is large: Vout 1 1 1 . 2 2 Vs C R 2 L 1 C LC 2 C L If is small: Vout Vs C 1 1 C 2 C 1. C EVALUATE: When is large, X C is small and X L is large so Z is large and the current is small. Both factors in 31.51. VC IX C are small. When is small, X C is large and the voltage amplitude across the capacitor is much larger than the voltage amplitudes across the resistor and the inductor. IDENTIFY: I V / Z and Pav 12 I 2 R. Z R 2 ( L 1/ C )2 V V EXECUTE: (a) I . 2 2 Z R L 1 C SET UP: 31-16 Chapter 31 2 1 1 V V 2R 2 (b) Pav I 2 R R 2 . 2 2 2Z R L 1 C (c) The average power and the current amplitude are both greatest when the denominator is smallest, which occurs 1 1 for 0 L , so 0 . 0C LC 25 2 100 V 200 2 . 2 2 2 200 2.00 H 1 [ (0.500 106 F)] 40 ,000 2 2 2 2 ,000 ,000 2 (d) Pav The graph of Pav versus is sketched in Figure 31.51. EVALUATE: Note that as the angular frequency goes to zero, the power and current are zero, just as they are when the angular frequency goes to infinity. This graph exhibits the same strongly peaked nature as the light purple curve in Figure 31.19 in the textbook. Figure 31.51 31.52. IDENTIFY: I VL I L and VC . C SET UP: Problem 31.51 shows that I EXECUTE: (a) VL I L (b) VC V R ( L 1/[ C ]) 2 2 . V L V L . 2 Z R 2 L 1/[C ] I I 1 . 2 C CZ C R L 1 [C ]2 (c) The graphs are given in Figure 31.52. EVALUATE: (d) When the angular frequency is zero, the inductor has zero voltage while the capacitor has voltage of 100 V (equal to the total source voltage). At very high frequencies, the capacitor voltage goes to zero, 1 while the inductor’s voltage goes to 100 V. At resonance, 0 1000 rad s, the two voltages are equal, and LC are a maximum, 1000 V. Figure 31.52 NAlternating Current 31.53. 31-17 U B 12 Li 2 . U E 12 Cv 2 . IDENTIFY: SET UP: Let x denote the average value of the quantity x. i 2 12 I 2 and vC2 12 VC2 . Problem 31.51 shows that I V R ( L 1/[ C ]) 2 2 . Problem 31.52 shows that VC V C R ( L 1/[C ]) 2 2 . 2 I 1 2 2 12 L EXECUTE: (a) U B 12 Li 2 U B 12 L i 2 12 LI rms 4 LI . 2 2 1 V U E 12 CvC2 U E C vC2 12 CVC2,rms 12 C C 14 CVC2 2 2 (b) Using Problem 31.51a UB 1 1 V2 LI 2 L 4 4 R 2 L 1 C 2 2 LV 2 . 4 R 2 L 1 C 2 2 1 1 V V2 Using Problem (31.47b): U E CVC 2 C . 2 4 4 2C 2 R2 L 1 C 2 4 2C R2 L 1 C (c) The graphs of the magnetic and electric energies are given in Figure 31.53. EVALUATE: (d) When the angular frequency is zero, the magnetic energy stored in the inductor is zero, while the electric energy in the capacitor is U E CV 2 4. As the frequency goes to infinity, the energy noted in both inductor and capacitor go to zero. The energies equal each other at the resonant frequency where 0 UB UE 1 and LC LV 2 . 4R2 Figure 31.53 31.54. IDENTIFY: At any instant of time the same rules apply to the parallel ac circuit as to parallel dc circuit: the voltages are the same and the currents add. SET UP: For a resistor the current and voltage in phase. For an inductor the voltage leads the current by 90° and for a capacitor the voltage lags the current by 90°. EXECUTE: (a) The parallel L-R-C circuit must have equal potential drops over the capacitor, inductor and resistor, so vR vL vC v. Also, the sum of currents entering any junction must equal the current leaving the junction. Therefore, the sum of the currents in the branches must equal the current through the source: i iR iL iC . (b) iR v is always in phase with the voltage. iL v lags the voltage by 90, and iC vC leads the voltage L R by 90. The phase diagram is sketched in Figure 31.54. (c) From the diagram, I I R I C I L 2 2 2 (d) From part (c): I V 2 V V V C . R L 2 2 1 1 1 V C . But I , so 2 Z R L Z 2 1 1 C . 2 R L 31-18 Chapter 31 EVALUATE: For large , Z 1 . The current in the capacitor branch is much larger than the current in the C other branches. For small , Z L. The current in the inductive branch is much larger than the current in the other branches. Figure 31.54 31.55. IDENTIFY: Apply the expression for I from problem 31.54 when 0 1/ LC . 2 1 1 C 2 R L 1 1 V 0C I C V 0C I so I I R and I is a minimum. EXECUTE: (a) At resonance, 0 0 L 0 L L LC SET UP: From Problem 31.54, I V 2 Vrms V2 at resonance where R < Z so power is a maximum. cos Z R (c) At 0 , I and V are in phase, so the phase angle is zero, which is the same as a series resonance. (b) Pav EVALUATE: (d) The parallel circuit is sketched in Figure 31.55. At resonance, iC iL and at any instant of time these two currents are in opposite directions. Therefore, the net current between a and b is always zero. (e) If the inductor and capacitor each have some resistance, and these resistances aren’t the same, then it is no longer true that iC iL 0 and the statement in part (d) isn’t valid. Figure 31.55 31.56. IDENTIFY: Refer to the results and the phasor diagram in Problem 31.54. The source voltage is applied across each parallel branch. SET UP: V 2Vrms 311 V EXECUTE: (a) I R V 311 V 0.778 A. R 400 (b) I C V C 311 V 360 rad s 6.00 106 F 0.672 A. I 0.672 A (c) arctan C arctan 40.8. I 0.778 A R (d) I I R2 IC2 31.57. 0.778 A 0.672 A 2 2 1.03 A. (e) Leads since 0. EVALUATE: The phasor diagram shows that the current in the capacitor always leads the source voltage. IDENTIFY and SET UP: Refer to the results and the phasor diagram in Problem 31.54. The source voltage is applied across each parallel branch. V V EXECUTE: (a) I R ; IC V C; I L . R L (b) The graph of each current versus is given in Figure 31.57a. (c) 0 : IC 0; I L . : IC ; I L 0. At low frequencies, the current is not changing much so the inductor’s back-emf doesn’t “resist.” This allows the current to pass fairly freely. However, the current in the capacitor goes to zero because it tends to “fill up” over the slow period, making it less effective at passing charge. At high frequency, the induced emf in the inductor resists the violent changes and passes little current. The capacitor never gets a chance to fill up so passes charge freely. NAlternating Current (d) 31-19 1 1 1000 rad sec and f 159 Hz. The phasor diagram is sketched in LC (2.0 H)(0.50 106 F) Figure 31.57b. 2 2 V V (e) I V C . R L 2 2 100 V 100 V 1 6 I (100 V)(1000 s )(0.50 10 F) 0.50 A 1 (1000 s )(2.0 H) 200 (f ) At resonance I L I C V C (100 V)(1000 s 1 )(0.50 106 F) 0.0500 A and I R V 100 V 0.50 A. R 200 EVALUATE: At resonance iC iL 0 at all times and the current through the source equals the current through the resistor. Figure 31.57 31.58. IDENTIFY: The average power depends on the phase angle . 2 1 SET UP: The average power is Pav = VrmsIrmscos , and the impedance is Z R 2 L . C , so = π/3 = 60°. tan = (XL – XC)/R, which gives tan 60° = (L – 1/C)/R. Using R = 75.0 Ω, L = 5.00 mH, and C = 2.50 µF and solving for we get = 28760 rad/s = 28,800 rad/s. (b) Z R 2 ( X L X C ) 2 , where XL = L = (28,760 rad/s)(5.00 mH) = 144 Ω and EXECUTE: (a) Pav = VrmsIrmscos = 1 2 (VrmsIrms), which gives cos = 1 2 XC = 1/C = 1/[(28,760 rad/s)(2.50 µF)] = 13.9 Ω, giving Z (75 )2 (144 13.9 ) 2 = 150 Ω; 31.59. I = V/Z = (15.0 V)/(150 Ω) = 0.100 A and Pav = 12 VI cos = 12 (15.0 V)(0.100 A)(1/2) = 0.375 W. EVALUATE: All this power is dissipated in the resistor because the average power delivered to the inductor and capacitor is zero. 2 R from Exercise 31.27 to calculate IDENTIFY: We know R, X C and so Eq.(31.24) tells us X L . Use Pav I rms Then calculate Z and use Eq.(31.26) to calculate for the source. I rms . Vrms SET UP: Source voltage lags current so 54.0. X C 350 , R 180 , Pav 140 W X XC EXECUTE: (a) tan L R X L R tan X C (180 ) tan(54.0) 350 248 350 102 2 R (Exercise 31.27). I rms (b) Pav Vrms I rms cos I rms Pav 140 W 0.882 A R 180 (c) Z R 2 ( X L X C ) 2 (180 ) 2 (102 350 ) 2 306 Vrms I rms Z (0.882 A)(306 ) 270 V. EVALUATE: We could also use Eq.(31.31): Pav Vrms I rms cos Pav 140 W Vrms 270 V, which agrees. The source voltage lags the current when I rms cos (0.882 A)cos(54.0) X C X L , and this agrees with what we found. 31-20 Chapter 31 31.60. IDENTIFY and SET UP: Calculate Z and I V / Z . EXECUTE: (a) For 800 rad s: Z R2 ( L 1 C )2 (500 )2 ((800 rad/s)(2.0 H) 1 ((800 rad/s)(5.0 107 F)))2 . Z 1030 . V 100 V 1 0.0971 A I 0.0971 A. VR IR (0.0971 A)(500 ) 48.6 V, VC 243 V C (800 rad s)(5.0 107 F) Z 1030 L 1 (C ) and VL IL (0.0971 A)(800 rad s)(2.00 H) 155 V. arctan 60.9. The graph of each R voltage versus time is given in Figure 31.60a. (b) Repeating exactly the same calculations as above for 1000 rad/s: Z R 500 ; 0; I 0.200 A; VR V 100 V; VC VL 400 V. The graph of each voltage versus time is given in Figure 31.60b. (c) Repeating exactly the same calculations as part (a) for 1250 rad/s: Z R 1030 ; 60.9; I 0.0971 A; VR 48.6 V; VC 155 V; VL 243 V. The graph of each voltage versus time is given in Figure 31.60c. 1 1 EVALUATE: The resonance frequency is 0 1000 rad/s. For 0 the phase LC (2.00 H)(0.500 F) angle is negative and for 0 the phase angle is positive. Figure 31.60 NAlternating Current 31.61. 31-21 IDENTIFY and SET UP: Eq.(31.19) allows us to calculate I and then Eq.(31.22) gives Z. Solve Eq.(31.21) for L. V 360 V 0.750 A EXECUTE: (a) VC IX C so I C X C 480 V 120 V 160 I 0.750 A (c) Z 2 R 2 ( X L X C ) 2 (b) V IZ so Z X L X C Z 2 R 2 , so X L X C Z 2 R 2 480 ± (160 )2 (80.0 )2 480 ± 139 X L 619 or 341 1 and X L L. At resonance, X C X L . As the frequency is lowered below the C resonance frequency X C increases and X L decreases. Therefore, for 0 , X L X C . So for X L 341 the XC (d) EVALUATE: angular frequency is less than the resonance angular frequency. is greater than 0 when X L 619 . But at these two values of X L , the magnitude of X L X C is the same so Z and I are the same. In one case ( X L 691 ) 31.62. the source voltage leads the current and in the other ( X L 341 ) the source voltage lags the current. IDENTIFY and SET UP: The maximum possible current amplitude occurs at the resonance angular frequency because the impedance is then smallest. EXECUTE: (a) At the resonance angular frequency 0 1/ LC , the current is a maximum and Z = R, giving Imax = V/R. At the required frequency, I = Imax/3. I = V/Z = Imax/3 = (V/R)/3, which means that Z = 3R. Squaring gives R2 + (L – 1/C)2 = 9R2 . Solving for gives = 3.192 105 rad/s and = 8.35 104 rad/s. I V 49.5 V 0.132 A. (b) V 2Vrms 2(35.0 V) 49.5 V. I max 3 3R 3(125 ) For 8.35 104 rad/s: R 125 and VR IR 16.5 ; X L L 125 and VL 16.5 V; 1 479 and VC 63.2 V. C For 3.192 105 rad/s: R 125 and VR IR 16.5 ; X L L 479 and VL 63.2 V; XC XC 1 C 125 and VC 16.5 V. For the lower frequency, X C X L and VC VL . For the higher frequency, X L X C and VL VC . EVALUATE: 31.63. IDENTIFY and SET UP: Consider the cycle of the repeating current that lies between t1 / 2 and t2 3 / 2. In this interval i 2I 0 (t ). I av t2 t2 1 1 2 i dt and I rms i 2dt t2 t1 t1 t2 t1 t1 3 / 2 I av EXECUTE: t2 1 1 3 / 2 2 I 0 2I 1 i dt (t ) dt 20 t 2 t t / 2 1 t2 t1 2 /2 2 3 2 2 2 I 2 I 9 I av 20 (2 I 0 ) 18 (9 12 1 4) 0 (13 13) 0. 8 2 8 2 4 2 I rms ( I 2 )av 2 I rms 4 I 02 3 t2 1 1 3 / 2 4I 02 i 2dt (t )2 dt t t2 t1 1 /2 2 3 / 2 /2 (t ) 2dt 4 I 02 3 3 / 2 1 ( t ) / 2 3 3 2 2 3 3 4 I 02 3 3 I 02 [1 1] 13 I 02 6 I 2 I rms I rms 0 . 3 EVALUATE: In each cycle the current has as much negative value as positive value and its average is zero. i 2 is always positive and its average is not zero. The relation between I rms and the current amplitude for this current is different from that for a sinusoidal current (Eq.31.4). 2 I rms 31-22 Chapter 31 31.64. IDENTIFY: Apply Vrms I rms Z 1 and Z R 2 ( X L X C ) 2 . LC 1 1 786 rad s. EXECUTE: (a) 0 LC (1.80 H)(9.00 107 F) 0 SET UP: (b) Z R 2 (L 1 C )2 . Z (300 )2 ((786 rad s)(1.80 H) 1 ((786 rad s)(9.00 107 F)))2 300 . I rms-0 Vrms 60 V 0.200 A. Z 300 (c) We want I 2 L2 1 V I rms-0 rms 2 Z Vrms R ( L 1 C ) 2 2 . R 2 ( L 1 C )2 2 4Vrms . 2 I rms-0 2 1 2 L 4Vrms 1 2L 4V 2 2 2 0. R 2 2 rms 0 and ( 2 ) 2 L2 2 R 2 2 C I C C C I rms-0 rms-0 2 Substituting in the values for this problem, the equation becomes ( 2 )2 (3.24) 2 (4.27 106 ) 1.23 1012 0. Solving this quadratic equation in 2 we find 2 8.90 105 rad 2 s 2 or 4.28 105 rad 2 /s 2 and 943 rad s or 654 rad s. (d) (i) R 300 , I rms-0 0.200 A, 1 2 289 rad s. (ii) R 30 , I rms-0 2A, 1 2 28 rad/s. (iii) R 3 , I rms-0 20 A, 1 2 2.88 rad/s. 31.65. EVALUATE: The width gets smaller as R gets smaller; I rms-0 gets larger as R gets smaller. IDENTIFY: The resonance frequency, the reactances, and the impedance all depend on the values of the circuit elements. SET UP: The resonance frequency is 0 1/ LC , the reactances are XL = L and XC = 1/C, and the impedance is Z R 2 ( X L X C ) 2 . 1 1/ 2, so 0 decreases by 12 . 2 L 2C (b) Since XL = L, if L is doubled, XL increases by a factor of 2. (c) Since XC = 1/C, doubling C decreases XC by a factor of 12 . EXECUTE: (a) 0 1/ LC becomes (d) Z R 2 ( X L X C ) 2 Z (2 R) 2 (2 X L 12 X C ) 2 , so Z does not change by a simple factor of 2 or 31.66. 1 2 . EVALUATE: The impedance does not change by a simple factor, even though the other quantities do. V N IDENTIFY: A transformer transforms voltages according to 2 2 . The effective resistance of a secondary V1 N1 R . circuit of resistance R is Reff ( N 2 / N1 ) 2 SET UP: N 2 275 and V1 25.0 V. EXECUTE: (a) V2 V1 ( N2 / N1 ) (25.0 V)(834/ 275) 75.8 V R 125 13.6 ( N 2 / N1 ) 2 (834 / 275) 2 EVALUATE: The voltage across the secondary is greater than the voltage across the primary since N 2 N1. The effective load resistance of the secondary is less than the resistance R connected across the secondary. 1 1 IDENTIFY: The resonance angular frequency is 0 and the resonance frequency is f 0 . LC 2 LC SET UP: 0 is independent of R. (b) Reff 31.67. EXECUTE: (a) 0 (or f 0 ) depends only on L and C so change these quantities. 31.68. (b) To double 0 , decrease L and C by multiplying each of them by 12 . EVALUATE: Increasing L and C decreases the resonance frequency; decreasing L and C increases the resonance frequency. IDENTIFY: At resonance, Z R. I V / R. VR IR, VC IX C and VL IX L . U E 12 CVC2 and U L 12 LI 2 . SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities. NAlternating Current EXECUTE: (a) I (c) VL IX L 31.69. R0C V L . R C V V L 0 L . R R C 1 1 V2 L 1 V2 (d) U C CVC2 C 2 L 2 . 2 2 R C 2 R 1 1 V2 (e) U L LI 2 L 2 . 2 2 R EVALUATE: At resonance VC VL and the maximum energy stored in the inductor equals the maximum energy stored in the capacitor. IDENTIFY: I V / R. VR IR, VC IX C and VL IX L . U E 12 CVC2 and U L 12 LI 2 . SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities. EXECUTE: (a) I V Z 0 2 . V L R 2 0 2 / 0C 2 (b) VC IX C 2 0C 0 L V R2 9 L 4C 2 V . 9 L R 4C 2 L C 2V R2 9 L 4C V . L V 2 . C 9 L 9 L 2 2 R R 4C 4C 1 2 LV 2 2 . (d) U C CVC 9 L 2 R2 4C 1 2 1 LV 2 . (e) U L LI 2 2 R2 9 L 4C EVALUATE: For 0 , VC VL and the maximum energy stored in the capacitor is greater than the maximum energy stored in the inductor. IDENTIFY: I V / R . VR IR , VC IX C and VL IX L . U E 12 CVC2 and U L 12 LI 2 . SET UP: The amplitudes of each time dependent quantity correspond to the maximum values of those quantities. EXECUTE: 20 . V V V (a) I . 2 2 Z 9 L R (20 L 1/ 20C ) 2 R 4C 1 V L V 2 . (b) VC IX C 2 0C C 9 L 9 L R2 R2 4C 4C (c) VL IX L 31.70. V V V 1 and I max . . At resonance L 2 2 Z C R R L 1/ C V (b) VC IX C 31-23 2 (c) VL IX L 2 0 L V 9 L R2 4C 2 1 LV (d) U C CVC2 . 2 9 L 2 8 R 4C L C 2V 9 L R 4C 2 . 31-24 Chapter 31 31.71. LV 2 . 9 L 2 R2 4C EVALUATE: For 0 , VL VC and the maximum energy stored in the inductor is greater than the maximum energy stored in the capacitor. IDENTIFY and SET UP: Assume the angular frequency of the source and the resistance R of the resistor are known. V I L L EXECUTE: Connect the source, capacitor, resistor, and inductor in series. Measure VR and VL . L VR IR R and L can be calculated. EVALUATE: There are a number of other approaches. The frequency could be varied until VC VL , and then this 31.72. frequency is equal to 1/ LC . If C is known, then L can be calculated. V IDENTIFY: Pav Vrms I rms cos and I rms rms . Calculate Z. R Z cos. Z SET UP: f 50.0 Hz and 2 f . The power factor is cos. 1 (e) U L LI 2 2 2 Vrms V 2 cos (120 V)2 (0.560) cos . Z rms 36.7 . Z Pav (220 W) R Z cos (36.7 )(0.560) 20.6 . EXECUTE: (a) Pav (b) Z R 2 X L 2 X L Z 2 R 2 (36.7 )2 (20.6 ) 2 30.4 . But 0 is at resonance, so the inductive and capacitive reactances equal each other. Therefore we need to add X C 30.4 . X C C 1 X C 31.73. IDENTIFY: SET UP: therefore gives 1 1 1.05 104 F. 2 f X C 2 (50.0 Hz)(30.4 ) (c) At resonance, Pav EVALUATE: resonance. 1 C V 2 (120 V)2 699 W. R 20.6 2 Pav I rms R and I rms is maximum at resonance, so the power drawn from the line is maximum at pR i 2 R. pL iL i I cost di q . pC i. dt C 1 EXECUTE: (a) pR i 2 R I 2 cos2 (t ) R VR I cos 2 (t ) VR I (1 cos(2t )). 2 1 T VR I T VR I T 1 Pav ( R) pR dt (1 cos(2t ))dt [t ]0 2 VR I . T 0 2T 0 2T T di (b) pL Li LI 2 cos(t )sin(t ) 12 VL I sin(2t ). But sin(2t )dt 0 Pav ( L) 0. 0 dt T q (c) pC i vCi VC I sin(t )cos(t ) 12 VC I sin(2t ). But sin(2t )dt 0 Pav (C ) 0. 0 C (d) p pR pL pc VR I cos 2 (t ) 12 VL I sin(2t ) 12 VC I sin(2t ) and V V VR and sin L C , so V V p VI cos(t )(cos cos(t ) sin sin(t )), at any instant of time. EVALUATE: At an instant of time the energy stored in the capacitor and inductor can be changing, but there is no net consumption of electrical energy in these components. dVL dVC IDENTIFY: VL IX L . 0 at the where VL is a maximum. VC IX C . 0 at the where VC is a d d maximum. V . SET UP: Problem 31.51 shows that I 2 R ( L 1/ C ) 2 p I cos(t )(VR cos(t ) VL sin(t ) VC sin(t )). But cos 31.74. EXECUTE: (a) VR maximum when VC VL 0 1 . LC NAlternating Current 31-25 dVL d V L dVL . 0 0.Therefore: d d R 2 ( L 1 C )2 d VL V 2 L( L 1 2C )( L 1 2C ) 2 2 0 . R ( L 1 C ) 2 ( L2 1 4C 2 ) . 2 2 3 2 2 2 ( R ( L 1 C ) ) R ( L 1 C ) (b) VL maximum when R2 1 R 2C 2 1 2L 1 and LC . 2 2C 2 C 2C 2 2 1 LC R 2C 2 2 . dVC d V dVC . 0 0. Therefore: d d d C R 2 ( L 1 C )2 V V ( L 1 2C )( L 1 2C ) 2 2 2 2 4 2 0 . R ( L 1 C ) ( L 1 C ). 2 2 3 2 2C R 2 ( L 1 C )2 C ( R ( L 1 C ) ) (c) VC maximum when 1 R2 2L 2. 2 L2 and LC 2 L C 2L 2 2 2 2 2 R L L . C EVALUATE: VL is maximum at a frequency greater than the resonance frequency and X C is a maximum at a frequency less than the resonance frequency. These frequencies depend on R , as well as on L and on C. IDENTIFY: Follow the steps specified in the problem. SET UP: In part (a) use Eq.(31.23) to calculate Z and then I V / Z . is given by Eq.(31.24). In part (b) let Z R iX . EXECUTE: (a) From the current phasors we know that Z R 2 ( L 1 C )2 . R2 2 L2 31.75. 2 1 Z (400 ) (1000 rad s)(0.50 H) 500 . 6 (1000 rad s)(1.25 10 F) V 200 V I 0.400 A. Z 500 2 (1000 rad s)(0.500 H) 1 (1000 rad s)(1.25 10 6 F) L 1 (C ) arctan (b) arctan . 36.9 400 R 1 1 (c) Zcpx R i L . Z cpx 400 i (1000 rad s)(0.50 H) (1000 rad s)(1.25 106 F) 400 300 i. C Z (400 )2 (300 ) 2 500 . V 200 V 8 6i 8 6i 8 6i A (0.320 A) (0.240 A)i. I 25 25 0.400 A. Z cpx (400 300i) 25 Im( I cpx ) 6 25 0.75 36.9. (e) tan Re( I cpx ) 8 25 (d) I cpx 8 6i (f ) VRcpx I cpx R (400 ) (128 96i )V. 25 8 6i VLcpx iI cpx L i (1000 rad s)(0.500 H) ( 120 160i) V. 25 1 8 6i i (192 256i) V. C 25 (1000 rad s)(1.25 106 F) VRcpx VLcpx VCcpx (128 96i) V (120 160i)V (192 256i) V 200 V. VCcpx i (g) Vcpx I cpx EVALUATE: Both approaches yield the same value for I and for .