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Chapter 21 Alternating Current Circuits and Electromagnetic Waves Problem Solutions 21.1 (a) VR,rms I rms R (b) VR,max 2 (c) I max (d ) Pav 21.2 (a) VR,rms 2I rms I 2 rms 8.00 A 2 R (c) Because R 2 2 Vrms Pav 136 V 11.3 A 12.0 768 W 2 1.20 102 V VR,rms 2 Vrms R 96.0 V 2 96.0 V 2 8.00 A 2 I rms R VR,max (b) Pav 8.00 A 12.0 R 2 Vrms Pav 1.70 102 V 1.20 102 V 60.0 W 2 2.40 102 (see above), if the bulbs are d esigned to operate at the sam e voltage, the 1.00 102 W will have the lower resistance. 21.3 The m eters m easure the rm s values of potential d ifference and current. These are Vrms Vmax 2 100 V 2 70.7 V , and I rms Vrms R 70.7 V 24.0 2.95 A 317 318 21.4 21.5 CH APTER 21 All lam ps are connected in parallel w ith the voltage source, so Vrms 120 V for each lam p. Also, the current is Irms Pav Vrms and the resistance is R Vrms I rms . 150 W 120 V I1, rms I 2, rms I 3, rms 100 W 120 V 1.25 A and R1 0.833 A and R3 R2 120 V 0.833 A The total resistance (series connection) is Req R1 120 V 1.25 A 96.0 144 R2 8.20 10.4 18.6 , so the current in the circuit is Vrms Req I rms 15.0 V 18.6 0.806 A The pow er to the speaker is then Pav 21.6 2 I rms Rspeaker The general form of the generator voltage is (a) VR,max (c) VR ,rms (d ) I rms 170 V and VR ,max 170 V 2 VR ,rms R (e) I max (f) 2 Pav I rms R (b) 2I rms 2 v Vmax sin 60 rad s 2 rad 10.4 6.76 W t , so by inspection 30.0 Hz 120 V 2 120 V 20.0 f 2 0.806 A 6.00 A 2 6.00 A 6.00 A 20.0 8.49 A 720 W (g) The argum ent of the sine function has units of t rad s s radians . At t 0.005 0 s , the instantaneous current is i v R 170 V 20.0 sin 60 rad s 0.005 0 s 170 V 20.0 sin 0.94 rad 6.9 A A lternating Current Circuits and Electromagnetic W aves 21.7 1 , so its units are 2 fC XC 1 1 Sec Farad 21.8 Vrms = 2 XC Volt Amp Ohm Vrms 2 f C 60.0 Hz 2.20 10 6 C/V =0.141 A 141 mA (b) I max = 2 240 V 2 50.0 Hz 2.20 10 6 C/V =0.235 A 235 mA I rms = Vrms =2 f C XC I rms 2 C Vrms Vrms , so 0.30 A 2 1 2 fC XC (a) (b) I rms 21.11 2 Volt Coulomb Sec I max = 2 120 V 2 f 21.10 1 1 Sec Coulomb Volt I max = 2 I rms = (a) 21.9 319 4.0 10 6 4.0 10 2 Hz F 30 V 1 2 VC ,rms 60.0 Hz 12.0 10 36.0 V 221 XC 6 221 F 0.163 A (c) I max (d ) No. The current reaches its m axim um value one-quarter cycle before the voltage reaches its m axim um value. From the d efinition of capacitance, the capacitor reaches its m axim um charge w hen the voltage across it is also a m axim um . Consequently, the m axim um charge and the m axim um current d o not occur at the sam e tim e. I rms = so 2 I rms 2 0.163 A Vrms =2 f C XC C Vmax 2 I f Vmax fC 0.231 A Vmax 2 0.75 A 2 60 Hz 170 V 2 1.7 10 5 F 17 F 320 21.12 CH APTER 21 (a) By inspection, (b) Also by inspection, 80 I max 0.354 A (c) I rms = 2 0.500 A = 2 VC ,max (d ) X C 21.13 1 2 fC XC XL 2 f L , and from t I (a) XL L 2 fL 2 (b) I rms (c) f I max VL ,rms XL 2 I rms VC ,max VC ,rms rad s , so f 98.0 V 2 2 80 rad s 2 rad 2 69.3 V 40.0 Hz 196 1 XC 1 , so C C (e) XL 21.15 98.0 V 0.500 A I max are then L 21.14 98.0 V , so VC ,max 80 I t L 1 rad s 196 2.03 10 t , w e have L I 5 F 20.3 F . The units of self ind uctance Volt sec . The units of ind uctive reactance are given by amp 1 sec Volt sec Amp Volt Amp 80.0 Hz 25.0 10 3 H 78.0 V 12.6 Ohm 12.6 6.19 A 2 6.19 A 8.75 A The required ind uctive reactance is XL VL ,max VL ,max I max 2 I rms and the need ed ind uctance is L 21.16 Given: vL XL 2 f VL,max 2 2 f I rms 1.20 102 V sin 30 t 4.00 V 2 2 300.0 Hz 2.00 10 3 A and L 0.500 H 0.750 H A lternating Current Circuits and Electromagnetic W aves (a) By inspection, (c) XL (d ) I rms 1.20 102 V VL ,max 2 2 fL L VL ,rms XL 30 rad s 2 2 15.0 Hz VL,max 1.20 102 V , so that (b) Also by inspection, VL ,rms rad s , so f 30 84.9 V 2 30 rad s 0.500 H 84.9 V 47.1 47.1 1.80 A (e) I max (f) The phase d ifference betw een the voltage across an ind uctor and the current through the ind uctor is L 90 , so the average pow er d elivered to the ind uctor is 2 I rms 321 2 1.80 A PL,av I rms VL,rms cos L 2.55 A I rms VL,rms cos 90 0 (g) When a sinusoid al voltage w ith a peak value VL,max is applied to an ind uctor, the current through the ind uctor also varies sinusoid ally in tim e, w ith the sam e frequency as the applied voltage, and has a m axim um value of I max VL,max X L . H ow ever, the current lags behind the voltage in phase by a quarter-cycle or VL,max sin t , the current as a 2 radians . Thus, if the voltage is given by vL function of tim e is i I max sin through it w ill be i 2.55 A sin 30 t (h) When i 1.00 A , w e have: 30 t and t t 2 sin 1 2 . In the case of the given ind uctor, the current 2 . 1.00 A 2.55 A , or sin 30 t 2 1.00 A 2.55 A sin 1 s 20.9 ms 2 rad 0.403 rad 30 rad s 2.09 10 2 0.392 0.403 rad 322 21.17 CH APTER 21 From L N N B, total B B I (see Section 20.6 in the textbook), the total flux through the coil is L I w here B is the flux through a single turn on the coil. Thus, L I max B ,total max L 21.18 2 Vrms XL 2 v 80.0 V and R2 Z 2 120 V 2 fL (a) The applied voltage is Vmax Vmax XL L Vmax sin 2 80.0 V sin 150t , so w e have that t 150 rad s . The im ped ance for the circuit is 2 f XC 0.450 T m 2 60.0 Hz R2 2 fL 1 2 fC 2 R2 L 1 C 2 2 or Z 40.0 Vmax Z (b) I max 21.19 2 XC 1 2 fC Z R2 (a) I rms 150 rad s 80.0 10 80.0 V 57.5 XL 60.0 Hz 40.0 10 2 XC Vrms Z (b) VR, rms I rms R (c) VC , rms I rms X C (d ) tan 1 50.0 30.0 V 83.0 XL R 6 2 F 150 rad s 125 10 66.3 2 0 66.3 0.361 A 0.361 A 50.0 XC H 1.39 A 1 2 3 1 0.361 A 66.3 tan 1 0 66.3 50.0 18.1 V 23.9 V 53.0 so, the voltage lags behind the current by 53° 83.0 6 F 57.5 A lternating Current Circuits and Electromagnetic W aves 21.20 (a) XL 2 fL 2 XC 1 2 fC Z R2 Vmax 2 H XC 2 2 500 126 XC tan R 719 2 719 776 194 V 126 1 F 126 0.250 A 776 XL 6 50.0 Hz 4.43 10 XL 1 3 1 I max Z tan (b) 50.0 Hz 400 10 719 49.9 500 Thus, the current leads the voltage by 49.9 21.21 (a) XL 2 fL 2 XC 1 2 fC Z R2 21.22 1 2 240 Hz 0.250 10 XL Vmax Z (b) I max XC 2 XC 6 2 900 140 V 1.44 103 XL 1 3.77 103 240 Hz 2.50 H 0.097 2 A 1 3.77 2.65 tan (d ) 0 , so the voltage leads the current XL 2 fL 2 XC 1 2 fC Z R2 (a) VR,rms tan XL 60.0 Hz 0.100 H I rms R 2 50.0 51.2 37.7 60.0 Hz 10.0 10 XC 103 900 1 2 2 103 3.77 2.65 (c) R F 2.65 103 2 2.75 A 50.0 6 265 F 37.7 138 V 265 2 233 (e) 1.44 k 323 324 CH APTER 21 (b) VL,rms (c) VC I rms X C 2.75 A 265 729 V (d ) Vrms I rms Z 2.75 A 233 641 21.23 I rms X L XL 2 fL 2 XC 1 2 fC (a) 60.0 Hz 0.400 H Z LC 0 2 60.0 Hz 3.00 10 XL XL 2 XC XC 2 6 884 F 2 60.0 VRC , rms XL 0 XC 2 151 884 2 735 2 fL 2 60 Hz 2.8 H Z R2 XC (a) I max Vmax Z I max R VL, max I max X L 884 90.0 V 735 733 2 89.8 V 886 886 108 V 1.1 103 1.2 103 170 V 1.6 103 VR, max 2 Vrms Z RC Z RLC XL 2 733 90.0 V 735 60.0 I rms Z RC XL XC Vrms Z LC Z RLC I rms Z LC R2 (b) Z RC (b) 151 Vrms Z RLC I rms VLC , rms 21.24 104 V 1 R2 Z RLC and 2.75 A 37.7 2 1.1 103 0 2 0.11 A 0.11 A 1.2 103 0.11 A 1.1 103 1.3 102 V 1.2 102 V 1.6 103 A lternating Current Circuits and Electromagnetic W aves (c) When the instantaneous current is a m axim um i across the resistor is vR iR I max R 325 I max , the instantaneous voltage 1.3 102 V . The instantaneous VR, max voltage across an ind uctor is alw ays 90° or a quarter cycle out of phase w ith the instantaneous current. Thus, w hen i Imax , vL 0 . Kirchhoff’ s loop rule alw ays applies to the instantaneous voltages around a closed path. Thus, for this series circuit, vsource vR vL and at this instant w hen i Imax w e have vsource 1.3 102 V I max R 0 (d ) When the instantaneous current i is zero, the instantaneous voltage across the resistor is vR iR 0 . Again, the instantaneous voltage across an ind uctor is a quarter cycle out of phase w ith the current. Thus, w hen i 0 , w e m ust have vL 1.2 102 V . Then, applying Kirchhoff’ s loop rule to the VL, max instantaneous voltages around the series circuit at the instant w hen i 0 gives vsource vR XC 1 2 fC Z RC R2 21.25 and (a) XC (b) Z (c) I max 0 2 50.0 103 rms Vb, rms R2 12 2 5 000 V 1.33 108 I rms Rb 2 2 R2 1.20 102 V 107 X C2 1.12 A F 3.76 10 3.76 10 60.0 Hz 30.0 10 1.33 108 6 2 1.33 108 5 5 1.33 108 A A 50.0 103 1 0 XC Vmax Z 1.2 102 V 60.0 Hz 20.0 10 X C2 Z RC 1 2 fC VL, max 1 Vsecondary I rms Therefore, 21.26 vL F 60.0 1.88 V 88.4 2 88.4 2 107 326 CH APTER 21 (d ) The phase angle in this RC circuit is tan Since XL 1 XC tan R 1 0 88.4 60.0 55.8 0 , the voltage lags behind the current by 55.8° . Ad d ing an ind uctor w ill change the im ped ance and hence the current in the circuit. 21.27 (a) XL 2 f L 2 50.0 Hz 0.250 H (b) XC 1 2 fC (c) Z R2 21.28 1 1 2 XL Vmax Z (d ) I max 78.5 50.0 Hz 2.00 10 XC 2 150 2.10 102 V 1.52 103 XL XC 1 F 0.138 A 78.5 1.59 103 78.5 1590 1590 tan 2 1.52 103 1.52 k (f) VR,max I max R VL,max I max X L 0.138 A 78.5 VC ,max I max X C 0.138 A 1.59 103 84.3 150 0.138 A 150 1.59 k 138 mA (e) R tan 2 6 20.7 V 10.8 V 219 V The voltage across the resistor is a m axim um w hen the current is a m axim um , and the m axim um value of the current occurs w hen the argum ent of the sine function is 2. The voltage across the capacitor lags the current by 90° or 2 rad ians, w hich correspond s to an argum ent of 0 in the sine function and a voltage of vC the voltage across the ind uctor lead s the current by 90° or to an argum ent in the sine function of 0 . Sim ilarly, 2 rad ians, correspond ing rad ians or 180°, giving a voltage of vL 0 . A lternating Current Circuits and Electromagnetic W aves 21.29 and (a) 21.30 XL 2 f L 2 50.0 Hz 0.185 H XC 1 2 fC Z ad R2 I rms Vrms Z ad Zab 1 2 XC Vmax 2 2 41.0 Vrms XL 58.1 , and Vrms (c) Zcd XC 49.0 , and Vrms (d ) Zbd XL XC Z R2 (a) I max (b) , so 9.10 Vrms XL 60 Hz 2.5 10 XC Vmax Z 2 1.2 103 170 V 1.6 103 VR, max I max R VC , max I max X C 6 bd 2 41.0 2.59 A 104 V 2.59 A 58.1 I rms Zcd cd 49.0 2.59 A 40.0 I rms Zbc bc 1 2 2 I rms Zab ab 49.0 58.1 150 V Z ad , so F 2 40.0 (b) Zbc 1 2 fC 6 50.0 Hz 65.0 10 XL R 40.0 XC 58.1 2.59 A 49.0 I rms Zbd 2.59 A 9.10 1.1 103 F 2 0 1.1 103 2 0.11 A 0.11 A 1.2 103 0.11 A 1.1 103 1.3 102 V 1.2 102 V 1.6 103 150 V 127 V 23.6 V 327 328 CH APTER 21 (c) When the instantaneous current i is zero, the instantaneous voltage across the resistor is vR iR 0 . The instantaneous voltage across a capacitor is alw ays 90° or a quarter cycle out of phase w ith the instantaneous current. Thus, w hen i 0 , vC and qC C vC 1.2 102 V VC , max 2.5 10 6 F 1.2 102 V 3.0 10-4 C 300 C Kirchhoff’ s loop rule alw ays applies to the instantaneous voltages around a closed path. Thus, for this series circuit, vsource vR vC 0 , and at this instant w hen i 0 , w e have vsource 0 vC 1.2 102 V VC , max (d ) When the instantaneous current is a m axim um (i across the resistor is vR iR I max R VR, max I max ) , the instantaneous voltage 1.3 102 V . Again, the instantaneous voltage across a capacitor is a quarter cycle out of phase w ith the current. Thus, w hen i Imax , w e m ust have vC 0 and qC C vC 0 . Then, applying Kirchhoff’ s loop rule to the instantaneous voltages around the series circuit at the instant w hen i Imax and vC 0 gives vsource 21.31 Vrms I rms vC 0 104 V 0.500 A (a) Z (b) 2 Pav Irms R gives R (c) Z and 21.32 vR Given v R2 X L2 , so X L L Vmax sin XL 2 f t vsource vR VR , max 1.3 102 V 208 Pav 10.0 W 2 rms 0.500 A I Z2 204 2 60.0 Hz R2 40.0 2 208 2 40.0 2 204 0.541 H 90.0 V sin 350t , observe that 350 rad s . Also, the net reactance is X L XC Vmax 90.0 V and 2 fL 1 2 fC L 1 C. A lternating Current Circuits and Electromagnetic W aves (a) XL XC 1 C L so the im ped ance is (b) I rms Vrms Z 1 350 rad s 0.200 H Vmax Z R2 X L Z 2 350 rad s 25.0 10 XC 90.0 V 2 2 50.0 6 329 44.3 F 2 44.3 66.8 0.953 A 2 66.8 (c) The phase d ifference betw een the applied voltage and the current is tan 1 XL XC 44.3 50.0 1 tan R 41.5 so the average pow er d elivered to the circuit is Vmax P av I rms Vrms cos I rms cos 0.953 A 2 21.33 2 cos 41.5 45.4 W Please see the textbook for the statem ent of Problem 21.21 and the answ ers for that problem . There, you should find that Vmax 140 V, Imax 0.097 2 A, and 51.2 . The average pow er d elivered to the circuit is then Pav I rms Vrms cos 21.34 90.0 V I max Vmax 2 cos 2 0.097 2 A 140 V 2 cos 51.2 4.26 W The rm s current in the circuit is I rms Vrms Z 160 V 80.0 2.00 A and the average pow er d elivered to the circuit is Pav I rms 21.35 (a) Vrms cos I rms VR,rms 2 Pav I rms R I rms I rms R Thus, R VR , rms I rms I rms I rms R I rms 50 V 0.28 A 2 I rms R VR , rms , so I rms 1.8 102 2.00 A Pav VR , rms 2 22.0 14 W 50 V 88.0 W 0.28 A 330 CH APTER 21 R2 (b) Z X L2 , w hich yield s XL R2 L XL 2 f and 21.36 Vrms I rms Z2 2 2 90 V 0.28 A R2 2.7 102 2 60 Hz 1.8 102 600 Hz 6.0 10 3 H 2 fL 2 XC 1 2 fC 23 Z R2 (a) VR , rms I rms R VL , rms I rms X L Vrms XL Z 10 V 28 23 8.2 V VC , rms I rms X C Vrms XC Z 10 V 28 11 3.9 V No , VR, rms 1 XL 11 600 Hz 25 10-6 F 2 XC 25 2 Vrms R Z VL, rms 2.7 102 0.72 H XL 2 2 23 11 10 V 28 VC , rms 25 2 28 8.9 V 9.0 V 8.2 V 3.8 V 21 V 10 V (b) The power delivered to the resistor is the greatest. N o pow er losses occur in an id eal capacitor or ind uctor. (c) Pav 21.37 Vrms Z 2 I rms R 2 R 10 V 28 2 25 3.2 W (a) The frequency of the station should m atch the resonance frequency of the tuning circuit. At resonance, X L X C or 2 f0 L 1 2 f0C , w hich gives f02 1 4 2 LC or f0 1 2 1 LC 2 3.00 10 6 H 2.50 10 5.81 107 Hz 12 F 58.1 MHz A lternating Current Circuits and Electromagnetic W aves (b) 331 Yes, the resistance is not needed. The resonance frequency is found by sim ply equating the ind uctive reactance to the capacitive reactance, w hich lead s to Equation 21.19 as show n above. 21.38 (a) The resonance frequency of a RLC circuit is f0 1 2 1 4 f02C L (b) XL For f0 1 2 f0 LC f0 max 2.00 10 9.00 109 Hz 1.56 10 10 H 12 10 H 156 pH F 8.82 1 4 f 02 L , so C 2 2 5 5.00 10 Hz 2 6 1.60 10 Hz (a) At resonance, X L R2 (b) When X L 2.0 10 6 5.1 10 8 F 51 nF 4.9 10 9 F 4.9 nF H XL XC 1 2 2.0 10 2 R2 0 R 2 H 15 1 w hich yield s 2 fC 1 LC 6 X C so the im ped ance w ill be X C , w e have 2 fL 2 2 1 4 f 9.00 10 Hz 1.56 10 2 1600 kHz 1.60 106 Hz C Cmin Z 9 1 4 For f0 2 500 kHz 5.00 105 Hz min C Cmax 21.40 4 2 f0 L 2 f0 21.39 1 2 LC . Thus, the ind uctance is 0.20 H 75 10 41 Hz 6 F (c) The current is a m axim um at resonance w here the im ped ance has its m inim um value of Z R. 332 CH APTER 21 (d ) At f 60 Hz , X L and Z 0 Thus, 2 15 0 2 35 Vmax 75 2 150 V Z 1 LC 3 10.0 10 60 Hz 75 10 1000 rad s H 100 10 6 F 2 000 rad s L 2 000 rad s 10.0 10 3 H XC 1 C 1 I rms 2 2.5 A 2 43 XL Z 1 , XC 43 1 2 f0 2 60 Hz 0.20 H 75 Vrms Z Thus, I rms 21.41 2 R2 Vrms Z 2 000 rad s 100 10 XL 2 XC 50.0 V 18.0 The average pow er is Pav 10.0 6 20.0 5.00 F 2 20.0 5.00 2 2.78 A 2 I rms R 2 2.78 A 10.0 77.3 W and the energy converted in one period is E Pav T 21.42 Pav 2 77.3 The resonance frequency is Also, X L L and X C 1 C 0 J s 2 f0 2 2 000 rad s 1 LC 0.243 J 18.0 6 F 35 , A lternating Current Circuits and Electromagnetic W aves (a) At resonance, X C R2 02 Thus, Z 1 2 XL 2 I rms R 1 4 0 Z 3 750 so Pav (d ) At 2 0 Z 1500 so Pav 2 I rms R XC 2 2 L C L 30.0 3.00 H 3.00 10-6 F 120 V 30.0 30.0 500 2 1 XL 4 2 XL 30.0 250 0 120 V 3 750 2 30.0 2 000 0 , and I rms 0.080 0 A 120 V 1 500 2 30.0 1 000 4.00 A 480 W , XC 2 2 XC 500 2 000 0 2 000 2 1500 0.080 0 A 0.032 0 A ; XL Vrms Z 0 , and I rms 2 I rms R LC 4.00 A 0.080 0 A ; XL 1 L R , I rms 1 XL 2 ; XL 120 V 1 500 and I rms (c) At 0 R2 Z so Pav 0 2 Pav I rms R and (b) At XL 0.192 W , XC 4 XC 4 000 0 0.032 0 A 3.07 10 , XC 2 W 1 XC 2 0.080 0 A 0.192 W 30.7 mW 500 0 333 334 CH APTER 21 (e) At 4 0 ; XL 2 I rms R , XC 4 000 0 120 V 3 750 , and I rms Z 3 750 so Pav 4 XL 2 0.032 0 A 1 XC 4 250 0 0.032 0 A 30.0 3.07 10 2 W 30.7 mW The pow er d elivered to the circuit is a m axim um w hen the rm s current is a m axim um . This occurs w hen the frequency of the source is equal to the resonance frequency of the circuit. 21.43 The m axim um output voltage Vmax 2 is related to the m axim um input voltage N2 Vmax 1 , w here N1 and N 2 are the num ber of N1 turns on the prim ary coil and the second ary coil respectively. Thus, for the given transform er, Vmax 1 by the expression Vmax 2 1500 170 V 250 Vmax 2 1.02 103 V Vrms and the rm s voltage across the second ary is 21.44 Vmax 2 2 2 1.02 103 V 2 721 V . (a) The output voltage of the transform er is V2,rms N2 N1 V1,rms 1 120 V 13 (b) Assum ing an id eal transform er, Poutput 9.23 V Pinput , and the pow er d elivered to the CD player is Pav 2 Pav 1 I1,rms V1,rms 0.250 A 120 V 30.0 W A lternating Current Circuits and Electromagnetic W aves 21.45 335 The pow er input to the transform er is Pav V1, rms I1, rms input For an id eal transform er, Pav 3 600 V 50 A 1.8 105 W Pav V2, rms I 2, rms ouput input so the current in the long- d istance pow er line is Pav I 2, rms input V2, rms 1.8 105 W 1.8 A 100 000 V The pow er d issipated as heat in the line is then Plost I 2,2 rms Rline 1.8 A 2 3.2 102 W 100 The percentage of the pow er d elivered by the generator that is lost in the line is % Lost 21.46 Plost 100% Pinput 3.2 102 W 1.8 105 W 100% 0.18% (a) Since the transform er is to step the voltage down from 120 volts to 6.0 volts, the second ary m ust have fewer turns than the prim ary. (b) For an id eal transform er, Pav Pav input ouput or V1, rms I1, rms V2, rms I 2, rms so the current in the prim ary w ill be I1, rms V2, rms I 2, rms 6.0 V 500 mA V1, rms 120 V 25 mA (c) The ratio of the second ary to prim ary voltages is the sam e as the ratio of the num ber of turns on the second ary and prim ary coils, V2 V1 N2 N1 . Thus, the num ber of turns need ed on the second ary coil of this step d ow n transform er is N2 N1 V2 V1 400 6.0 V 120 V 20 turns 336 21.47 CH APTER 21 (a) At 90% efficiency, Pav Thus, if Pav 0.90 Pav output 1000 kW output the input pow er to the prim ary is Pav (b) I1, rms (c) 21.48 Rline I 2, rms Pav 1.1 103 kW V1, rms input V1, rms Pav 1000 kW V1, rms output V2, rms 4.50 10 4 transmitted Vrms Thus, Pav loss output 0.90 1 000 kW 0.90 1.1 103 kW 3.1 102 A 8.3 103 A 290 transmitted 5.00 106 W 500 103 V 2 I rms Rline input 1.0 106 W 120 V m 6.44 105 m Pav Pav 1.1 106 W 3 600 V (a) The pow er transm itted is Pav so I rms input 10.0 A 2 Vrms I rms 10.0 A 290 2.90 104 W 29.0 kW (b) The pow er input to the line is Pav input Pav transmitted Pav loss 5.00 106 W 2.90 104 W 5.03 106 W and the fraction of input pow er lost d uring transm ission is fraction Pav Pav loss input 2.90 104 W 5.03 106 W 0.005 77 or 0.577% A lternating Current Circuits and Electromagnetic W aves 337 (c) It is im possible to d eliver the need ed pow er w ith an input voltage of 4.50 kV. The m axim um line current w ith an input voltage of 4.50 kV occurs w hen the line is shorted out at the custom er’ s end , and th is current is I rms Vrms Rline max 4 500 V 15.5 A 290 The m axim um input pow er is then Pinput max Vrms I rms max 4.50 103 V 15.5 A 6.98 104 W 6.98 kW This is far short of m eeting the custom er’ s request, and all of this pow er is lost in the transm ission line. 21.49 From v f , the w avelength is 3.00 108 m s 75 Hz v f 4.00 106 m 4 000 km The required length of the antenna is then, L 21.50 (a) t d c 4 1000 km , or about 621 m iles. N ot very practical at all. 6.44 1018 m 1y 8 3.00 10 m s 3.156 107 s 6.80 102 y (b) From Table C.4 (in Append ix C of the textbook), the average Earth-Sun d istance is d 1.496 1011 m , giving the transit tim e as t d c 1.496 1011 m 1 min 3.00 108 m s 60 s 8.31 min (c) Also from Table C.4, the average Earth -Moon d istance is d tim e for the round trip as t 2d c 2 3.84 108 m 3.00 108 m s 2.56 s 3.84 108 m , giving the 338 21.51 CH APTER 21 The am plitud es of the electric and m agnetic com ponents of an electrom agnetic w ave are related by the expression Emax Bmax c , thus the am plitud e of the m agnetic field is Bmax 21.52 c 330 V m 3.00 108 m s 1 or 1.10 10 6 T 1 0 21.53 Emax c 0 c 4 10 7 N s2 C2 8.854 10 12 C2 N m 2 2.998 108 m s (a) The frequency of an electrom agnetic w ave is f c , w here c is the speed of light, and is the w avelength of the w ave. The frequencies of the tw o light sources are then c 3.00 108 m s f red 4.55 1014 Hz Red : -9 660 10 m red and c 3.00 108 m s 3.19 1014 Hz Infrared : f IR -9 940 10 m IR (b) The intensity of an electrom agnetic w ave is proportional to the square of its am plitud e. If 67% of the incid ent intensity of the red light is absorbed , then the intensity of the em erging w ave is 100% 67% 33% of the incid ent intensity, or If 0.33Ii . H ence, w e m ust have Emax, f If Emax, i Ii 0.33 0.57 339 A lternating Current Circuits and Electromagnetic W aves 21.54 If I0 is the incid ent intensity of a light beam , and I is the intensity of the beam after passing through length L of a fluid having concentration C of absorbing m olecules, the Beer-Lam bert law states that log10 I I0 is a constant. CL w here For 660-nm light, the absorbing m olecules are oxygenated hem oglobin. Thus, if 33% of this w avelength light is transm itted through blood , the concentration of oxygenated hem oglobin in the blood is log10 0.33 CHBO2 [1] L The absorbing m olecules for 940-nm light are d eoxygenated hem oglobin, so if 76% of this light is transm itted through the blood , the concentration of these m olecules in the blood is log10 0.76 CHB [2] L Divid ing equation [2] by equation [1] gives the ratio of d eoxygenated hem o globin to oxygenated hem oglobin in the blood as log10 0.76 CHB CHBO2 log10 0.33 0.25 or CHB 0.25CHBO2 Since all the hem oglobin in the blood is either oxygenated or d eoxygenated , it is necessary that CHB CHBO2 1.00 , and w e now have 0.25CHBO2 CHBO2 1.0 . The fraction of hem oglobin that is oxygenated in this blood is then 1.0 1.0 0.25 CHBO2 0.80 or 80% Som eone w ith only 80% oxygenated hem oglobin in the blood is probably in serious trouble need ing supplem ental oxygen im m ed iately. 21.55 From Intensity Emax Bmax E and max 2 0 Bmax c , w e find Intensity 2 c Bmax 2 0 Thus, Bmax and Emax 2 0 c Bmax c 2 4 Intensity 3.35 10 10 7 T mA 8 3.00 10 m s 6 T 3.00 108 m s 1340 W m2 1.01 103 V m 3.35 10 6 T 340 21.56 CH APTER 21 (a) To exert an upw ard force on the d isk, the laser beam should be aim ed vertically upw ard , striking the low er surface of the d isk. To just levitate the d isk, the upw ard force exerted on the d isk by the beam should equal the w eight of the d isk. The m om entum that electrom agnetic rad iation of intensity I, incid ent norm ally on a perfectly reflecting surface of area A , d elivers to that surface in tim e t is given by Equation 21.30 as p 2U c 2 IA t c . Thus, from the im pulse-m om entum theorem , the average force exerted on the reflecting surface is F p t 2IA c . Then, to just levitate the surface, F 2IA c mg and the required intensity of the incid ent rad iation is I (b) I mgc 2A mgc 2 r2 mgc 2 A . 5.00 10 3 kg 9.80 m s 2 3.00 108 m s 2 -2 4.00 10 m 2 1.46 109 W m2 (c) Propulsion by light pressure in a significant gravity field is im practica l because of the enorm ous pow er requirem ents. In ad d ition, no m aterial is perfectly reflecting, so the absorbed energy w ould m elt the reflecting surface. 21.57 The d istance betw een ad jacent antinod es in a stand ing w ave is Thus, 2 6.00 cm c 21.58 f 2 12.0 cm 0.120 m , and 0.120 m 2.45 109 Hz 2.94 108 m s At Earth’ s location, the w ave fronts of the solar rad iation are spheres w hose rad ius is the Sun-Earth d istance. Thus, from Intensity Pav 21.59 From 21.60 c f f Intensity 4 r 2 c , w e find f 3.00 108 m s 27.33 106 Hz c 1 340 W m2 Pav Pav A 4 r2 4 3.00 108 m s 5.50 10 7 m 11.0 m , the total pow er is 1.49 1011 m 5.45 1014 Hz 2 3.74 1026 W A lternating Current Circuits and Electromagnetic W aves 21.61 (a) For the AM band , c min f max c max f min 3.00 108 m s 1 600 103 Hz 188 m 3.00 108 m s 540 103 Hz 556 m (b) For the FM band , c min f max c max 21.62 f min 3.00 108 m s 108 106 Hz 2.78 m 3.00 108 m s 88 106 Hz 3.4 m The transit tim e for the rad io w ave is tR dR c 100 103 m 3.00 108 m s 3.33 10 4 s 0.333 ms and that for the sound w ave is ts ds vsound 3.0 m 343 m s 8.7 10 3 s 8.7 ms Thus, the radio listeners hear the news 8.4 ms before the studio audience because rad io w aves travel so m uch faster than sound w aves. 341 342 21.63 CH APTER 21 If an object of m ass m is attached to a spring of spring constant k, the natural frequency of vibration of that system is f k m 2 . Thus, the resonance frequency of the C=O d ouble bond w ill be 1 2 f k 1 2 moxygen 2 800 N m 2.66 10 26 kg 5.2 1013 Hz atom and the light w ith this frequency has w avelength 3.00 108 m s 5.2 1013 Hz c f 5.8 10 6 m 5.8 m The infrared region of the electrom agnetic spectrum ranges from 1 mm d ow n to 700 nm 0.7 m . Thus, the required wavelength falls within the infrared region . min 21.64 max Since the space station and the ship are m oving tow ard one another, the frequency after being Doppler shifted is fO f S 1 u c , so the change in frequency is f fO fS fS u c 6.0 1014 Hz 1.8 105 m s 3.0 108 m s 3.6 1011 Hz and the frequency observed on the spaceship is fO 21.65 fS 6.0 1014 Hz 3.6 1011 Hz f 6.003 6 1014 Hz Since you and the car ahead of you are m oving aw ay from each other (getting farther apart) at a rate of u 120 km h 80 km h 40 km h , the Doppler shifted frequency you w ill d etect is fO f fO fS f S 1 u c , and the change in frequency is fS u c 4.3 1014 Hz 40 km h 3.0 108 m s 0.278 m s 1 km h The frequency you w ill d etect w ill be fO fS f 4.3 1014 Hz 1.6 107 Hz 4.299 999 84 1014 Hz 1.6 107 Hz A lternating Current Circuits and Electromagnetic W aves 21.66 The d river w as d riving tow ard the w arning lights, so the correct form of the Doppler shift equation is fO f S 1 u c . The frequency em itted by the yellow w arning light is c fS S 3.00 108 m s 580 10 9 m 5.17 1014 Hz and the frequency the d river claim s that she observed is c fO O 3.00 108 m s 560 10 9 m 5.36 1014 Hz The speed w ith w hich she w ould have to approach the light for the Doppler effect to yield this claim ed shift is u 21.67 (a) At f XC c fO fS 3.00 108 m s 1 5.36 1014 Hz 1 5.17 1014 Hz 60.0 Hz , the reactance of a 15.0- F capacitor is 1 2 fC 1 2 60.0 Hz 15.0 10 6 F 177 and the im ped ance of this RC circuit is Z (b) I rms R2 Vrms Z X C2 50.0 1.20 102 V 184 2 177 0.652 A 2 184 652 mA 1.1 107 m s 343 344 CH APTER 21 (c) After ad d ition of an ind uctor in series w ith the resistor and capacitor, the R2 im ped ance of the circuit is Z XL 2 XC 2 50.0 2 fL 177 2 . To red uce the current to one-half the value found above, the im ped ance of the circuit m ust be d oubled to a value of Z 2 184 368 . Thus, 2 50.0 or 2 2 fL 177 2 fL 177 368 2 2 368 2 50.0 177 365 Since the ind uctance cannot be negative, the potential solution associated w ith the low er sign m ust be d iscard ed , leaving L 21.68 177 365 2 60.0 Hz 1.44 H Suppose you cover a 1.7 m -by-0.3 m section of beach blanket. Suppose the elevation angle of the Sun is 60°. Then the target area you fill in the Sun’ s field of view is 1.7 m 0.3 m cos30 0.4 m2 . The intensity the rad iation at Earth’ s surface is Isurface absorbed . Since I E Pav E A A 0.5 I surface A t t , the absorbed energy is 0.5 0.6 I incoming A t 0.5 0.6 1 340 W m 2 0.4 m 2 3 600 s 21.69 Z R2 200 Thus, Pav and cost I 2 rms 2 XC 2 2 Vrms Z R E rate 8.9 10 R2 3 2 fC Pav R 6 105 J or ~106 J 2 60 Hz 5.0 10 2 0.6 Iincoming and only 50% of this is 120 V 5.7 102 6 2 F 5.7 102 2 200 8.9 W 8.9 10 t rate kW 24 h 8.0 cents kWh 1.7 cents 3 kW A lternating Current Circuits and Electromagnetic W aves 21.70 L , so XL 1 C Then, X C L From 0 XL L 1 w hich gives C XL L X L XC C 1 2 f0 LC 12 C or L 8.0 2 96 [1] C 1 , w e obtain LC 2 f0 2 2 Substituting from Equation [1], this becom es 96 1 C2 2 2 f0 or 1 C 2 f0 1 2 96 2.6 10 2 2 2 000 Hz 96 F 2.5 10 3 2.5 mH 5 F 26 F Then, from Equation [1], L 21.71 V R Thus, L 2.6 10 5 H 12.0 V 19.0 0.630 A DC I DC Z 21.72 2 96 R2 2 fL 2 Z 2 R2 2 f Vrms I rms 24.0 V 0.570 A 2 42.1 2 19.0 42.1 2 9.97 10 60.0 Hz 2 H 99.7 mH 3.00 108 m s 1.0 1010 Hz . Therefore, the 2 3.00 10 m 1 1.0 1010 Hz , giving resonance frequency of the circuit is f 0 2 LC (a) The required frequency is f C c 1 2 f0 1 2 L 2 10 10 Hz 2 6.3 10 400 10 12 H 13 F 0.63 pF 345 346 CH APTER 21 2 A 0 (b) C 0 d , so d (c) XC (a) Emax Bmax c , so Bmax Emax c XL (c) Pav Intensity (a) (b) Vrms I rms Z R VDC I DC From Z L A 17 12 V 2.0 A Z 2 R2 2 f 6 10 H 8.4 mm 25 T V m 6.7 10 7 m 16 T T mA 5.3 10 d2 4 Intensity 20.0 m 2 16 12 3 2 1.7 10 4 14 W 2 H 6.0 12 V 3.0 A R2 6.7 10 2 4 W m 8.4 10 C N m 0.20 10 Emax Bmax 2 0 m 2 0.20 10 6 V m 3.00 108 m s 5.3 10 21.74 12 1.0 1010 Hz 400 10 2 f0 L 2 (b) Intensity 3 F 1.0 10 8.85 10 0 21.73 13 6.3 10 C d 4.0 X L2 R2 6.0 2 2 fL 2 4.0 60 Hz 2 , w e find 2 1.2 10 12 mH 17 W m2 A lternating Current Circuits and Electromagnetic W aves 21.75 347 (a) From Equation 21.30, the m om entum im parted in tim e t to a perfectly reflecting sail of area A by norm ally incid ent rad iation of intensity I is p 2U c 2 IA t c . From the im pulse-m om entum theorem , the average force exerted on the sail is then Fav Fav m (b) aav (c) From t 21.76 2 IA t c p t t 0.536 N 6 000 kg 2 8.93 10 1 2 at , w ith v0 2 x v0 t 2 1340 W m2 6.00 104 m2 2IA c 5 m s2 0 , the tim e is 2 3.84 108 m x aav 8.93 10 5 0.536 N 3.00 108 m s ms 2 2.93 106 s 1d 8.64 104 s 33.9 d (a) The intensity of rad iation at d istance r from a point source, w hich rad iates total pow er P , is I P A P 4 r 2 . Thus, at d istance r 2.0 in from a cell phone rad iating a total pow er of P 2.0 W 2.0 103 mW , the intensity is 2.0 103 mW I 4 2.0 in 2.54 cm 1 in 2 6.2 mW cm2 This intensity is 24% higher than the maximum allowed leakage from a microwave at this d istance of 2.0 inches. (b) If w hen using a Blue tooth head set (em itting 2.5 m W of pow er) in the ear at d istance rh 2.0 in 5.1 cm from center of the brain, the cell phone (em itting 2.0 W of pow er) is located in the pocket at d istance rp 1.0 m 1.0 102 cm from the brain, the total rad iation intensity at the brain is I total I phone 2.0 103 mW I headset 4 or I total 1.6 10 2 1.0 10 cm 2 mW cm 2 2.5 mW 2 7.6 10 4 3 5.1 cm mW cm 2 2 1.6 10 2.4 10 2 2 mW cm 2 mW 7.6 10 cm2 3 mW cm2 0.024 mW cm 2