Alternating Current Circuits and Electromagnetic Waves

Transcription

1 hapter Alternating urrent ircuits and Electromagnetic Wes Quick Quizzes. (a, (c. The erage power is proportional to the current which is non-zero even though the erage current is zero. (a is only valid for an open circuit, for which R. (b and (d can never be true because i for A currents.. (b. hoices (a and (c are incorrect because the unaligned sine curves in Figure.9 mean the voltages are out of phase, and so we cannot simply add the imum (or voltages across the elements. (n other words, VR + V + V even though v vr + v + v.. (b. Note that this is a D circuit. However, changing the amount of iron inside the solenoid changes the magnetic field strength in that region and results in a changing magnetic flux through the loops of the solenoid. This changing flux will generate a back emf that opposes the current in the circuit and decreases the brightness of the bulb. The effect will be present only while the rod is in motion. f the rod is held stationary at any position, the back emf will disappear, and the bulb will return to its original brightness. 4. (b, (c. The radiation pressure (a does not change because pressure is force per unit area. n (b, the smaller disk absorbs less radiation, resulting in a smaller force. For the same reason, the momentum in (c is reduced. 5. (b, (d. The frequency and welength of light wes are related by the equation λ f c or f c λ, where c is the speed of light is a constant within a given medium. Thus, the frequency and welength are inversely proportional to each other, when one increases the other must decrease.

2 HAPTER Answers to Even Numbered onceptual Questions. At resonance, X X. This means that the impedance Z R + ( X X reduces to Z R. 4. The purpose of the iron core is to increase the flux and to provide a pathway in which nearly all the flux through one coil is led through the other. 6. The fundamental source of an electromagnetic we is a moving charge. For example, in a transmitting antenna of a radio station, charges are caused to move up and down at the frequency of the radio station. These moving charges set up electric and magnetic fields, the electromagnetic we, in the space around the antenna.. Energy moves. No matter moves. You could say that electric and magnetic fields move, but it is nicer to say that the fields stay at that point and oscillate. The fields vary in time, like sports fans in the grandstand when the crowd does the we. The fields constitute the medium for the we, and energy moves.. The erage value of an alternating current is zero because its direction is positive as often as it is negative, and its time erage is zero. The erage value of the square of the current is not zero, however, since the square of positive and negative values are always positive and cannot cancel.. The brightest portion of your face shows where you radiate the most. Your nostrils and the openings of your ear canals are particularly bright. Brighter still are the pupils of your eyes. 4. No, the only element that dissipates energy in an A circuit is a resistor. nductors and capacitors store energy during one half of a cycle and release that energy during the other half of the cycle, so they dissipate no net energy. 6. The changing magnetic field of the solenoid induces eddy currents in the conducting core. This is accompanied by Rconversion of electrically-transmitted energy into internal energy in the conductor.. The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagrams throughout the chapter. Kirchhoff s loop rule is true at any instant, but the voltages across different circuit elements are not simultaneously at their imum values. Do not forget that an inductor can induce an emf in itself and that the voltage across it is 9 ahead of the current in the circuit in phase.. nsulation and safety limit the voltage of a transmission line. For an underground cable, the thickness and dielectric strength of the insulation between the conductors determines the imum voltage that can be applied, just as with a capacitor. For an overhead line on towers, the designer must consider electrical breakdown of the surrounding air, possible accidents, sparking across the insulating supports, ozone production, and inducing voltages in cars, fences, and the roof gutters of nearby houses. Nuisance effects include noise, electrical noise, and a prankster lighting a hand-held fluorescent tube under the line.

3 Alternating urrent ircuits and Electromagnetic Wes Answers to Even Numbered Problems. (a 9 Ω (b 45 Ω 4.,,.5 A, R R 96. Ω,,. A, R 44 Ω 6. (a 6 V (b 6. Hz (c (d. A. (a 4 ma (b 5 ma. ma. 4 ma 4..6 A 6. > 7. H. (a 94 V (b current leads by (a V (b 4 V (c 79 V (d 64 V (e. (a. A (b (c (d V V, V V R,, v, v v V, q µ R R source source v v V, v, q 4. (a. A (b (c (d vr vsource V, v, v V R R,, v V V, V V v source 6. (a nf or 4 nf (b 5.5 kv. (a.49, 4.5 W (b.44,.7 W. (a W,.6 (b 56 W,.79 DV 4 V DV DV 65 V DVR V 77.6 DV 64 V DV 79 V. (a VR, + V, + V, V V, but accounting for phases and adding the voltages vectorially does yield V. (b The power loss delivered to the resistor. No power losses occur in an ideal capacitor or inductor. (c. W 4. (a Z R 5 Ω (b 4 Hz (c At resonance (d.5 A 6. (a 4 W (b.9 W (c.7 mw (d.9 W (e.7 mw Maximum power is delivered at resonance frequency.

4 4 HAPTER. (a turns (b.6 W 4. (a Fewer turns (b 5 ma (c turns 4. (a 9. kw (b.5% (c The imum power that can be input to the line at 4.5 kv is far less than 5. MW, and it is all lost in the transmission line m s 46. % W 5.. m 5. Radio listeners hear the news.4 ms before the studio audience because radio wes trel much faster than sound wes Hz, the frequency increases by.6 Hz 7. m s 6 ~ J 6..5 mh, 6 µ F 6. (a.6 pf (b.5 mm (c 5 Ω 64. (a 6. Ω (b mh X R c

5 Alternating urrent ircuits and Electromagnetic Wes 5 Problem Solutions. (a V ( V V 4 V (b R V 5. Ω. A (c R 4 V 5. Ω. A or. A. A (d P R. A 5. Ω. W. kw. P ( R R R R R, so R ( P (a f P 75. W, then (b f P W, then ( 7 V R 9 Ω 75. W ( 7 V ( W R 45 Ω. The meters measure the values of potential difference and current. These are V V 7.7 V, and 7.7 V R 4. Ω.95 A.4 All lamps are connected in parallel with the voltage source, so V V for each lamp. Also, the current is and the resistance is R V. 5 W V.5 A and R R 96. Ω V.5 A,, W. A V, and V R 44 Ω. A

6 6 HAPTER.5 The total resistance (series connection is R R + R. Ω+.4 Ω.6 Ω, so the current in the circuit is 5. V.6 A Req.6 Ω eq The power to the speaker is then P R speaker.6 A.4 Ω 6.76 W.6 (a V 5 V, so 5 V 6 V V (b ω 77 rad s f π π 6. Hz (c At t ( s, v ( π 5 V sin 77 rad s s 5 V sin rad (d R 5 V 5. Ω. A.7 X π f, so its units are Volt Volt Ohm Sec Farad Sec oulomb Volt oulomb Sec Amp. ( V π f ( X (a π V 6. Hz. /V.4 A 4 ma (b π 4 V 5. Hz. /V.5 A 5 ma.9 π f( X, so f. A π 4. F V 4. Hz ( V π

7 Alternating urrent ircuits and Electromagnetic Wes 7. π f( X π 9. Hz.7 /V 4. V. A ma. π f π f ( V X so.75 A π f 6 Hz 7 V ( π 5.7 F 7 µ F X. ω 4 V rad s 6. F.4 A 4 ma or ( π. X π f, and from then [ ] [ ][ t] [ ] ε ε [ ] Volt sec Amp, we he ε ( t t. The units of self inductance are. The units of inductive reactance are given by Volt sec Volt X [ f][ ] Ohm sec Amp Amp.4 The imum current in the purely inductive circuit is 4 V X ω ( π rad s(. H.7 A so.7 A.6 A

8 HAPTER.5 The ratio of inductive reactance at f 5. Hz to that at f 6. Hz is ( ( X X π f f f 5. Hz 54. Ω 45. Ω, so ( X ( X π f f f 6. Hz The imum current at f 5. Hz is then ( V X X 45. Ω.4 A.6 The imum current in this inductive circuit will be ( X ( ( π f Thus, if <. ma, it is necessary that ( 5. V > or > π f. ma π. Hz. A ( 7. H N Φ.7 From flux through a single turn on the coil. Thus, B, the total flux through the coil is B, total ( B total Φ, X ( π f π V 6. Hz Φ NΦ where is the B.45 T m Φ B. (a X π f π X 5. Hz 4 H 6 Ω 79 Ω π f π 5. Hz 4.4 F ( 5. Hz 4 mh 5 W 4.4 mf Z R X X ( 5 ( Ω + Ω V Z.5 A 776 Ω 94 V Ω 776 Ω

9 Alternating urrent ircuits and Electromagnetic Wes 9 (b X X 6 Ω 79 Ω φ tan tan 49.9 R 5 Ω Thus, the current leads the voltage by X 66. Ω π f π 6. Hz 4. F ( Z R X X ( 5. ( Ω + Ω. Ω (a Z. V. Ω.6 A (b VR, R.6 A 5. Ω. V (c V, X.6 A 66. Ω.9 V (d X X 6. Ω φ tan tan 5. R 5. Ω so, the voltage lags behind the current by Hz. H 7.7 Ω. X π f π X π f π 6. Hz. F Z R X X ( 65 Ω ( 5. ( Ω + Ω (a Ω Ω VR, R.75 A 5. Ω V (e DV 4 V DVR V (b V, X.75 A 7.7 Ω 4 V (c V, X.75 A 65 Ω 79 V (d V Z.75 A Ω 64 V DV DV 65 V 77.6 DV 64 V DV 79 V

10 HAPTER X π f π 4 Hz.5 H.. (a Ω X.7 π f π 4 Hz.5 F ( Ω Z R + X X 9 Ω +..7 Ω.4 kω (b 4 V Z.4 Ω. A (c..7 X X φ tan tan Ω 5 R 9 Ω (d φ >, so the voltage leads the current. X. π f π 6 Hz.5 F ( Ω Z R + X X. Ω +. Ω.6 (a 7 V Z.6 Ω. A (b V R Ω R,. A.. V V X Ω,. A.. V (c When the instantaneous current i is zero, the instantaneous voltage across the resistor is v ir. The instantaneous voltage across a capacitor is always 9 R or a quarter cycle out of phase with the instantaneous current. Thus, when i, Ω v V,. V and q v µ F. V. Kirchhoff s loop rule always applies to the instantaneous voltages around a closed path. Thus, for this series circuit, v v + v and at this instant when i, we he v + V V source, source R

11 Alternating urrent ircuits and Electromagnetic Wes (d When the instantaneous current is a imum ( i, the instantaneous voltage across the resistor is v ir R R R,. V current. Thus, when i, we must he. Again, the instantaneous voltage across a capacitor is a quarter cycle out of phase with the v and q ( v. Then, applying Kirchhoff s loop rule to the instantaneous voltages around the series circuit at the instant when i gives vsource v R + v V R, +. V. X π f π X 6. Hz.4 H 5 Ω π f π 6. Hz. F ( R 4 Ω ( 6. ( 5 Z R + X X Ω + Ω 4 Ω 76 Ω and Z R (a Z ( X + X X X 7 Ω 9. V V, Z Z ( 7 Ω 9.6 V ZR 76 Ω (b Z R ( X R + 6. Ω + 4 Ω 6 Ω 9. V VR, ZR ZR ( 6 Ω V ZR 76 Ω.4 X π f π 6 Hz. H. Ω Z R + X X. Ω +. Ω.6 7 V (a. A Z.6 Ω Ω

12 HAPTER (b V R Ω R,. A.. V V X Ω,. A.. V (c When the instantaneous current is a imum ( i, the instantaneous voltage across the resistor is v ir R R R,. V instantaneous current. Thus, when i, v. The instantaneous voltage across an inductor is always 9 or a quarter cycle out of phase with the. Kirchhoff s loop rule always applies to the instantaneous voltages around a closed path. Thus, for this series circuit, vsource vr + v and at this instant when i we he v R+. source V (d When the instantaneous current i is zero, the instantaneous voltage across the resistor is v ir. Again, the instantaneous voltage across an inductor is a R quarter cycle out of phase with the current. Thus, when i, we must he v. Then, applying Kirchhoff s loop rule to the V,. V instantaneous voltages around the series circuit at the instant when i gives v v + v + V source R,. V.5 X. π f π 6. Hz. F ( Ω Z R + X 5. Ω +. Ω. Ω R and ( sec ondary ZR. 5 V Ω 5.76 A 5 Therefore, Vb Rb,.76 A 5. Ω. V

13 Alternating urrent ircuits and Electromagnetic Wes.6 (a X π f π Z Thus, 4 Hz.5 H.9 Ω Ω V A.5 H V Hz 5. W ( 5. ( 5. ± ± Ω Ω X X Z R ± 5.7 Ω and X X ± 5.7 Ω or π f.9 4 Ω ± 5.7 Ω This yields π Hz.9 Ω 4 ( ± 5.7 Ω nf or 4 nf (b ( V Z R + X ( 4. A ( 5. Ω + (.9 4 Ω coil coil V 5.5 kv Notice that this is a very large voltage!.7 X π f π X 5. Hz.5 H 5. Ω π f π 5. Hz 65. F ( ad 49. Ω ( 4. ( Z R + X X Ω + Ω Ω 4. Ω and ( 5 V Zad Zad ( 4. Ω.55 A (a Z R 4. Ω ab, so ( V Z ab.55 A 4. Ω V ab (b Z X 5. Ω, and ( V Z bc bc.55 A 5. Ω 5 V (c Z X 49. Ω, and ( V Z cd bc cd.55 A 49. Ω 7 V cd

14 4 HAPTER (d Z X X 9.5, so bd Ω ( V Z bd.55 A 9.5 Ω.6 V bd. (a X.4 Ω π f π 6. Hz. F ( + 5. Ω +.4 Ω Ω Z R X Z V.94 A Ω X.4 tan Ω tan 6.5 φ R 5. Ω and power factor cosφ cos P ( V cos φ ( V(.94 A( W (b X π f π 6. Hz. H Ω + 5. Ω + Ω 4 Ω Z R X Z V.9 A 4 Ω X tan Ω tan 66. φ R 5. Ω and power factor cosφ cos( P ( V φ ( cos V.9 A.44.7 W 4 V.9 (a Z Ω.5 A (b P R gives P. W R 4. Ω (.5 A

15 Alternating urrent ircuits and Electromagnetic Wes 5 (c Z R, so X Z R + X Ω 4. Ω 4 Ω and X 4 Ω π f 6. Hz π.54 H. (a X π f π X 6. Hz. H 7.7 Ω π f π 6. Hz F. Ω Z R X X P -6 ( (. ( Ω + Ω V Ω.6 Ω R R R. Ω W Z.6 Ω P R R. Ω and power factor cosφ R.6 Z.6 Ω (b The same calculations as shown in Part (a above, with f 5. Hz, give X.4 Ω, X 5.9 Ω, Z 5. Ω, P 56 W and power factor.79 4 W, so. A 5 V. (a R ( R ( R, R, Thus, R, 5 V R. Ω. A (b Z R, which yields + X 9 V X Z R R. Ω.7. A Ω and X.7 Ω π f 6 Hz π.7 H

16 6 HAPTER. X π f π X 6 Hz 6. H Ω π f π 6 Hz 5 F Ω Z R X X -6 ( ( 5 ( + Ω + Ω (a V R R R, Ω Ω V 5 Ω 9. V Z Ω V V, X X ( Ω. V Z Ω V V, X X ( Ω. V Z Ω No, V + V + V 9. V +. V +. V V V R,,, However, observe that if we take phases into account and add these voltages vectorially, we find ( VR ( V V, +,, 9. V +. V. V V (b The p ower delivered to the resistor is the greatest. No power losses occur in an ideal capacitor or inductor. V 5 Ω. W Z Ω (c P R R. The resonance frequency of the circuit should match the broadcast frequency of the station. f gives, π 4π f or µ 6 4π.9 Hz.4 F (.9 H.9 H

17 Alternating urrent ircuits and Electromagnetic Wes 7.4 (a At resonance, X X so the impedance will be (b When X ( X X Z R + R + R 5 Ω X, we he π f π f which yields f π π. H 75 F ( 4 Hz (c The current is a imum at resonance where the impedance has its minimum value of Z R. (d At f 6, Hz X π 6 Hz. H 75 Ω 5 Ω, π 6 Hz 75 F, X 6 ( and Z ( 5 Ω + 75 Ω 5 Ω 4 Ω Thus, Z ( 5 V Z ( Ω 4.5 A.5 f, so π 4π f For 5 f f min 5 kh z 5. Hz 5 4π 5. Hz. ( 6 For f f 6 khz.6 Hz min 6 4π.6 Hz. ( H ( ( H 5. F 5 nf F 4.9 nf.6 The resonance frequency is ω π f Also, X ω and X ω

18 HAPTER. H (a At resonance, X X ω Ω -6. F Thus, and Z R + R, P R V 4. A Z. Ω 4. A. Ω 4 W (b At ω ω ; X ( X ω 5 Ω, ( Z R X X X X ω Ω +. Ω + 5 Ω Ω 5 Ω and V. A 5 Ω so P R (c At ω ω 4. A. Ω.9 W ; X ( X ω 5 Ω, ( 4 V Z 75Ω, and. A 75 Ω so P R (d At ω ω X 4 X ω 4 Ω. A. Ω.7 W.7 mw ; X ( X ω Ω, X ( X ω 5 Ω V Z 5 Ω, and. A 5 Ω so P R. A. Ω.9 W

19 Alternating urrent ircuits and Electromagnetic Wes 9 (e At ω 4ω ; X 4( X ω 4 Ω, X ( X ω V Z 75Ω, and. A 75 Ω so P R 5 Ω 4. A. Ω.7 W.7 mw The power delivered to the circuit is a imum when the current is a imum. This occurs when the frequency of the source is equal to the resonance frequency of the circuit..7 ω π f rad s (. H( F Thus, ω ω rad s X X ω rad s. H. Ω 5. Ω ω Z R X X ( rad s( 6 F +. Ω +. Ω 5. Ω. Ω Z 5. V.77 A. Ω The erage power is P R.77 A. Ω 76.9 W and the energy converted in one period is π J π E P T P J ω s rad s N. (a V, ( V, N 9. V 4 turns turns V, V, so N N

20 HAPTER (b For an ideal transformer, ( P ( P ( Thus,,, input ouput V P 9. V.4 A.6 W input.9 The power input to the transformer is Pinput V 5,, 6 V 5 A. W For an ideal transformer, ( P ( V P so the current in the longdistance power line is P, lost ( P input (, ouput,, input 5. W. A V The power dissipated as heat in the line is then, R. A Ω. W line The percentage of the power delivered by the generator that is lost in the line is Plost. W % ost % %.% 5 Pinput. W.4 (a Since the transformer is to step the voltage down from volts to 6. volts, the secondary must he fewer turns than the primary. (b For an ideal transformer, current in the primary will be P P input or ouput ( V,, ( V,, so the, (,, ( 6. V( 5 ma, V 5 ma (c The ratio of the secondary to primary voltages is the same as the ratio of the number of turns on the secondary and primary coils, V N N. Thus, the number of turns needed on the secondary coil of this step down transformer is N 6. V ( 4 turns N V

21 Alternating urrent ircuits and Electromagnetic Wes.4 (a At 9% efficiency, ( P.9( P output input Thus, if ( P output kw the input power to the primary is ( P input ( P output kw. kw.9.9 (b (c ( P 6 input,. kw. W. A,, 6 V ( P 6 output, kw. W. A,, V R line 4.5 Ω m 6.44 m 9 Ω (a The power transmitted is ( P ( transmitted V so ( P 6 transmitted 5. W. A V 5 V Thus, ( P. A 9 Ω.9 4 R W 9. kw loss line (b The power input to the line is 6 4 ( P ( P ( P W+.9 W5. W input transmitted loss and the fraction of input power lost during transmission is ( P ( P loss fraction 5. W input 4.9 W 6.5 or.5%

22 HAPTER (c t is impossible to deliver the needed power with an input voltage of 4.5 kv. The imum line current with an input voltage of 4.5 kv occurs when the line is shorted out at the customer s end, and this current is ( 4 5 V 5.5 A R 9 Ω line The imum input power is then ( Pinput ( ( ( 4.5 V( 5.5 A W 6.9 kw This is far short of meeting the customer s request, and all of this power is lost in the transmission line..4 From v λ f, the welength is. m s v λ f 75 Hz 6 4. m 4 km The required length of the antenna is then, λ 4 km, or about 6 miles. Not very practical at all..44 c µ π ( 7 4 N s (.54 N m or c.99 m s.45 (a The frequency of an electromagnetic we is f c λ, where c is the speed of light, and λ is the welength of the we. The frequencies of the two light sources are then c. m s 4 Red: fred 4.55 Hz -9 λred 66 m and c. m s 4 nfrared: fr.9 Hz -9 λ 94 m R

23 Alternating urrent ircuits and Electromagnetic Wes (b The intensity of an electromagnetic we is proportional to the square of its amplitude. f 67% of the incident intensity of the red light is absorbed, then the % 67% % of the incident intensity, or intensity of the emerging we is.. Hence, we must he f i E, f E, i f..57 i.46 f is the incident intensity of a light beam, and is the intensity of the beam after passing through length of a fluid hing concentration of absorbing molecules, the Beer-ambert law states that log εwhere ε is a constant. ( For 66-nm light, the absorbing molecules are oxygenated hemoglobin. Thus, if % of this welength light is transmitted through blood, the concentration of oxygenated hemoglobin in the blood is HBO ε log. [] The absorbing molecules for 94-nm light are deoxygenated hemoglobin, so if 76% of this light is transmitted through the blood, the concentration of these molecules in the blood is HB ( ε log.76 [] Dividing equation [] by equation [] gives the ratio of deoxygenated hemoglobin to oxygenated hemoglobin in the blood as HB log.76 log. HBO.5 or.5 HB HBO Since all the hemoglobin in the blood is either oxygenated or deoxygenated, it is necessary that HB + HBO., and we now he.5 HBO + HBO.. The fraction of hemoglobin that is oxygenated in this blood is then. HBO. or %. +.5 Someone with only % oxygenated hemoglobin in the blood is probably in serious trouble needing supplemental oxygen immediately.

24 4 HAPTER.47 The distance between adjacent antinodes in a standing we is λ λ 6. cm. cm. m, and Thus, (. m(.45 Hz c λ f.94 m s 9.4 At Earth s location, the we fronts of the solar radiation are spheres whose radius is the Sun-Earth distance. Thus, from ntensity, the total power is A 4π r P W ( ntensity( 4πr 4 4 π(.49 m 6.74 W m EB E.49 From n tensity and µ B Thus, c, we find cb ntensity µ B 7 ( π µ 4 T m A ( ntensity c. m s and E B c 6.5 T. m s. V m 4 W m.5 T.5. m s c λ 6 f 7. Hz. m.5 (a For the AM band, λ c. m s min f 6 Hz m λ c. m s fmin 54 Hz 556 m

25 Alternating urrent ircuits and Electromagnetic Wes 5 (b For the FM band, λ c. m s min 6 f Hz.7 m λ c. m s 6 fmin Hz.4 m.5 The transit time for the radio we is t R dr c. m s m 4. s. ms and that for the sound we is ds ts v 4 m s sound. m.7 s.7 ms Thus, the radio listeners hear the news.4 ms before the studio audience because radio wes trel so much faster than sound wes..5 f an object of mass m is attached to a spring of spring constant k, the natural frequency of vibration of that system is f k m π. Thus, the resonance frequency of the double bond will be O f k N m π m π oxygen.66 kg atom 5. Hz and the light with this frequency has welength. m s c λ µ f 5. Hz 5. m 5. m The infrared region of the electromagnetic spectrum ranges from λ mm down to λ min 7 nm.7 µ m. Thus, the required welength falls within the infrar ed region.

26 6 HAPTER.54 Since the space station and the ship are moving toward one another, the frequency after f f + u c, so the change in frequency is being Doppler shifted is O S 5 u 4. m s f fo fs fs ( 6. Hz.6 Hz c. m s and the frequency observed on the spaceship is f f + f + O S Hz.6 Hz 6. 6 Hz.55 Since you and the car ahead of you are moving away from each other (getting farther apart at a rate of u km h km h 4 km h, the Doppler shifted frequency you will detect is f f ( u c O, and the change in frequency is S u 4 4 km h.7 m s 7 f fo fs fs ( 4. Hz.6 Hz c. m s km h The frequency you will detect will be f f f O S + 4. Hz. 6 Hz Hz.56 The driver was driving toward the warning lights, so the correct form of the Doppler f f + u c. The frequency emitted by the yellow warning light is shift equation is O S f S S. m s c 9 λ 5 m Hz and the frequency the driver claims that she observed is f O O. m s c 9 λ 56 m Hz The speed with which she would he to approach the light for the Doppler effect to yield this claimed shift is 4 f O 5.6 Hz 7 u c (. m s 4. m s fs 5.7 Hz

27 Alternating urrent ircuits and Electromagnetic Wes 7.57 ( D. V R 9. Ω.6 A D 4. V Z R + ( π f 4. Ω.57 A Thus, ( 4. Ω ( 9. Ω π Z R π f 6. Hz 9.96 H 99.6 mh.5 Suppose you cover a.7 m-by-. m section of beach blanket. Suppose the elevation angle of the Sun is 6. Then the target area you fill in the Sun s field of view is.7 m. m cos.4 m. The intensity the radiation at Earth s surface is surface.6 incoming and only 5% of this is P E t absorbed. Since A A, the absorbed energy is incoming A( t E.5surface A t W m.4 m 6 s 6 J ~ J.59 Z R + ( X R + ( π f 6 π Ω + 6 Hz 5. F 5.7 Ω V Ω.9 W.9 kw Z 5.7 Ω Thus, P R R and cost E ( rate P t ( rate.9 kw 4 h. cents kwh.7 cents

28 HAPTER.6 X ω, so ω X Then, X ω X which gives ( ( Ω(. Ω or ( 96 X X Ω ( From ω π f, we obtain ( π f Substituting from Equation (, this becomes 96 Ω ( π f or ( π f 96 Ω π( π Hz 96 Ω 5.6 F 6 µ F Then, from Equation (, Ω F.5 H.5 mh.6 (a The box cannot contain a capacitor since a steady D current cannot flow in a series circuit containing a capacitor. Since the A and D currents are different, even when a. V potential difference is used in both cases, the box must contain a reactive element. The conclusion is that the box must contain a resistor and inductor connected in series. (b D R Ω D. V. A Z 5 Ω. V. A Since + R ( π f Z R X +, we find 5 ( Ω ( Ω π Z R π f 6 Hz mh

29 Alternating urrent ircuits and Electromagnetic Wes 9 c. m s.6 (a The required frequency is f. Hz. Therefore, the λ. m resonance frequency of the circuit is f. Hz, giving π 6. F ( π f ( π Hz ( 4 H.6 pf (b A, so d d ( ( 6. F. m.5 m.5 mm d.5 N m (c X X ( π f π. Hz 4 H 5 Ω E.6 (a B c, so B E. V m T c. m s (b EB n tensity µ (. V m( 6.7 T 7 ( 4π T m A 5. W m 7 ntensity A ntensity π d 4 (c P π (. m 7 4 ( 5. W m.7 W 4 V.64 (a Z. A 6. Ω

30 HAPTER (b D R D V 4. Ω. A From ( π Z R + X R + f, we find ( 6. Ω ( 4. Ω π Z R π f 6 Hz. H mh.65 (a The radiation pressure on the perfectly reflecting sail is ( ntensity 4 ( Wm p.9 N m c. m s so the total force on the sail is 4 F p A.9 N m 6. m.56 N (b F.56 N a.9 m s m 6 kg 5 (c From x v t+ at, with v, the time is ( x.4 ( m 6 d t 5 (.9 s.9 d 4 a.9 m s.64 s.66 We know that (, Z (, Z N Z N,, Z,, Also, for an ideal transformer, (, rm ( V V s,,, which gives V,,,, Therefore, N Z N V Z,,, or N Z N V, Z, N N Z This gives N N Z N Z Ω, or N Z. Ω

31 Alternating urrent ircuits and Electromagnetic Wes.67 onsider a cylindrical volume with V. iter. m and cross-sectional area A. m The length of this one liter cylinder is V d A. m. m. m magine this cylinder placed at the top of Earth s atmosphere, with its length perpendicular to the incident we fronts. Then, all the energy in the one liter volume of sunlight will strike the atmosphere in the time required for sunlight to trel the length of the cylinder. This time is d t c. m s. m. s The energy passing through the. m area of the end of the cylinder in this time is ( E P t ntensity A t ( 4 W m (. m (. 9 s 4.47 J.6 The capacitance of a parallel-plate capacitor is e A d, and its reactance in an A circuit is X π f. Observe that reducing the plate separation to one-half of its original value will double the capacitance and reduce the capacitive reactance to onehalf the original value. The impedance of an R series circuit in which X R is Z R + ( R X. When the applied voltage is V, the current in the circuit is V Z V R + ( R X. f now the plate separation, and hence the capacitive reactance, is cut to one-half the original value, the new impedance is Z R ( R X + and the new current will be V Z R + R X. f it is observed that, then we must he R or + ( X 4 R + R X ( R X R + ( R X + R R

32 HAPTER Expanding the last result yields R + R RX + X 4R + 4R 4RX + X which reduces to 6R RX and yields X R