c h L 75 10

Transcription

1 hapter (a) All the energy in the circuit resies in the capacitor when it has its axiu charge. The current is then zero. f Q is the axiu charge on the capacitor, then the total energy is c c h h U Q F (b) When the capacitor is fully ischarge, the current is a axiu an all the energy resies in the inuctor. f is the axiu current, then U = / leas to c h 6 U J H (a) We recall the fact that the perio is the reciprocal of the frequency. t is helpful to refer also to Fig The values of t when plate A will again have axiu positive charge are ultiples of the perio: t n n A nt f b J. 3 g n 500. s, Hz where n = 1,, 3, 4,. The earliest tie is (n = 1) t 5.00s. (b) We note that it takes t 1 T for the charge on the other plate to reach its axiu positive value for the first tie (copare steps a an e in Fig. 31-1). This is when plate A acquires its ost negative charge. Fro that tie onwar, this situation will repeat once every perio. onsequently, n1 n1 1 1 t T ( n 1) T n 1 T n1.50 s, 3 f 10 Hz where n = 1,, 3, 4,. The earliest tie is (n = 1) t.50s. A A. (c) At t 1 4 T, the current an the agnetic fiel in the inuctor reach axiu values for the first tie (copare steps a an c in Fig. 31-1). ater this will repeat every halfperio (copare steps c an g in Fig. 31-1). Therefore, 1333

2 1334 HAPTER 31 T ( n1) T T t n1 n 11.5 s, 4 4 where n = 1,, 3, 4,. The earliest tie is (n = 1) t 1.5s. 3. (a) The perio is T = 4(1.50 s) = 6.00 s. (b) The frequency is the reciprocal of the perio: f 1 T Hz.. s (c) The agnetic energy oes not epen on the irection of the current (since U B i ), so this will occur after one-half of a perio, or 3.00 s. 4. We fin the capacitance fro U 1 Q : Q U c c h Jh F Accoring to U Q, the current aplitue is Q c Hhc Fh A. 6. (a) The angular frequency is k F x 8. 0N 13 c.0 10 hb050. kgg 89 ra s. (b) The perio is 1/f an f = /. Therefore, T s. ra s (c) Fro = () 1/, we obtain F. 89 ra s 50. H b g b g 7. THNK This proble explores the analogy between an oscillating syste an an oscillating ass spring syste. EXPRESS Table 31-1 provies a coparison of energies in the two systes. Fro the table, we see the following corresponences:

3 x q x q, k,, v i, t t 1 q 1 1 kx, v i. ANAYZE (a) The ass correspons to the inuctance, so = 1.5 kg. (b) The spring constant k correspons to the reciprocal of the capacitance, 1/. Since the total energy is given by U = Q /, where Q is the axiu charge on the capacitor an is the capacitance, we have an c c h h Q U J k 37 N / / N (c) The axiu isplaceent correspons to the axiu charge, so 4 x ax () The axiu spee v ax correspons to the axiu current. The axiu current is 6 Q Q A H F onsequently, v ax = /s. b gc h EARN The corresponences suggest that an oscillating syste is atheatically equivalent to an oscillating ass spring syste. The electrical echanical analogy can also be seen by coparing their angular frequencies of oscillation: k 1 (ass-spring syste), ( circuit) 8. We apply the loop rule to the entire circuit: i q i q total 1 1 R 1 j j R j j irj ir j j t j t with 1 1 j,, R Rj j j j j 3 F

4 1336 HAPTER 31 an we require total = 0. This is equivalent to the siple R circuit shown in Fig. 31-7(b). 9. The tie require is t = T/4, where the perio is given by T /. onsequently, 6 T H F t We fin the inuctance fro 1 f / f Hz F s H. 11. THNK The frequency of oscillation f in an circuit is relate to the inuctance an capacitance by f 1/. EXPRESS Since f 1/, the saller value of gives the larger value of f, while the larger value of gives the saller value of f. onsequently, fax 1/ in, an f in 1/. ax ANAYZE (a) The ratio of the axiu frequency to the iniu frequency is f f ax in ax in 365pF pF (b) An aitional capacitance is chosen so the esire ratio of the frequencies is 160. MHz r MHz Since the aitional capacitor is in parallel with the tuning capacitor, its capacitance as to that of the tuning capacitor. f is in picofaras (pf), then The solution for is 365pF 10pF. 96.

5 1337 b365pfg b 96g b10pfg 36pF b ġ (c) We solve f 1/ for. For the iniu frequency, = 365 pf + 36 pf = 401 pf an f = 0.54 MHz. Thus, the inuctance is bg f bg c Fhc Hzh H. EARN One coul also use the axiu frequency conition to solve for the inuctance of the coil in (). The capacitance is = 10 pf + 36 pf = 46 pf an f = 1.60 MHz, so H. 1 6 f 4610 F Hz 1. (a) Since the percentage of energy store in the electric fiel of the capacitor is (1 75.0%) 5.0%, then which leas to q/ Q U E q / 50%. U Q / (b) Fro we fin i/ U U B i / 750%,. / 13. (a) The charge (as a function of tie) is given by q Qsint, where Q is the axiu charge on the capacitor an is the angular frequency of oscillation. A sine function was chosen so that q = 0 at tie t = 0. The current (as a function of tie) is an at t = 0, it is = Q. Since 1/, q i Qcos t, t b g c hc h Q. 00A H F (b) The energy store in the capacitor is given by

6 1338 HAPTER 31 an its rate of change is U U t E E q Q t sin Q sint cost We use the trigonoetric ientity cost sint 1 sin t b g U E Q sin t. t b g to write this as The greatest rate of change occurs when sin(t) = 1 or t = / ra. This eans t H F s. (c) Substituting = /T an sin(t) = 1 into U E /t = (Q /) sin(t), we obtain U E Q Q. t T T ax c hc h so Now T H F s, U E t ax s.7010 F 66.7 W. We note that this is a positive result, inicating that the energy in the capacitor is inee increasing at t = T/ The capacitors 1 an can be use in four ifferent ways: (1) 1 only; () only; (3) 1 an in parallel; an (4) 1 an in series. (a) The sallest oscillation frequency is f H.0 10 F F Hz (b) The secon sallest oscillation frequency is.

7 f H F (c) The secon largest oscillation frequency is f H.0 10 F () The largest oscillation frequency is f Hz Hz F5.010 F / H.010 F5.010 F (a) The axiu charge is Q = V ax = ( F)(3.0 V) = (b) Fro U Q / we get Hz. Q c Hhc Fh A. (c) When the current is at a axiu, the agnetic energy is at a axiu also: c hc h U B,ax H A J. 16. The linear relationship between (the knob angle in egrees) an frequency f is f f 0 F f HG K J F HG f0 KJ where f 0 = 10 5 Hz. Since f = / = 1/, we are able to solve for in ters of : f 1 / with S units unerstoo. After ultiplying by 10 1 (to convert to picofaras), this is plotte next:

8 1340 HAPTER (a) After the switch is thrown to position b the circuit is an circuit. The angular frequency of oscillation is 1/. onsequently, f Hz H F 3 6 (b) When the switch is thrown, the capacitor is charge to V = 34.0 V an the current is zero. Thus, the axiu charge on the capacitor is The current aplitue is Q = V = (34.0 V)( F) = Q fq 75 Hz A. 18. (a) Fro V = X we fin = /V. The perio is then T = / = V/ = 46.1 s. (b) The axiu energy store in the capacitor is UE 1 1 ( F)(0.50 V) V J. (c) The axiu energy store in the inuctor is also UB /6.88 nj. () We apply Eq as V = (i/t) ax. We can substitute = V / (cobining what we foun in part (a) with Eq. 31-4) into Eq (as written above) an solve for (i/t) ax. Our result is 3 i V V ( A) 7 t V / V (.010 F)(0.50 V) ax A/s.

9 1341 (e) The erivative of U B = 1 i leas to U B t 1 sint cost sin t. Therefore, U B t ax ( V A)(0.50 V) W. 19. The loop rule, for just two evices in the loop, reuces to the stateent that the agnitue of the voltage across one of the ust equal the agnitue of the voltage across the other. onsier that the capacitor has charge q an a voltage (which we ll consier positive in this iscussion) V = q/. onsier at this oent that the current in the inuctor at this oent is irecte in such a way that the capacitor charge is increasing (so i = +q/t). Equation then prouces a positive result equal to the V across the capacitor: V = (i/t), an we interpret the fact that i/t > 0 in this iscussion to ean that (q/t)/t = q/t < 0 represents a eceleration of the charge-builup process on the capacitor (since it is approaching its axiu value of charge). n this way we can check the signs in Eq (which states q/ = q/t ) to ake sure we have ipleente the loop rule correctly (a) We use U Q / to solve for : 1 Q 1 Vax Vax V F H A (b) Since f = /, the frequency is f H F Hz. (c) Referring to Fig. 31-1, we see that the require tie is one-fourth of a perio (where the perio is the reciprocal of the frequency). onsequently, t T 4 4 f e 3 Hz j s. 1. (a) We copare this expression for the current with i = sin(t+ 0 ). Setting (t+) = 500t = /, we obtain t = s. (b) Since = 500 ra/s = () 1/,

10 134 HAPTER ra / s F b g c h H. (c) The energy is U c hb g H 160. A J.. For the first circuit = ( 1 1 ) 1/, an for the secon one = ( ) 1/. When the two circuits are connecte in series, the new frequency is / / eq eq / where we use , 3. (a) The total energy U is the su of the energies in the inuctor an capacitor: A H J. q i U U E U B F (b) We solve U = Q / for the axiu charge: c hc h Q U F J (c) Fro U = /, we fin the axiu current: 6 c Jh 3 U H A. () f q 0 is the charge on the capacitor at tie t = 0, then q 0 = Q cos an cos F H G q Q KJ cos 1 1 F HG KJ For = the charge on the capacitor is ecreasing, for = 46.9 it is increasing. To check this, we calculate the erivative of q with respect to tie, evaluate for t = 0.

11 1343 We obtain Q sin, which we wish to be positive. Since sin(+46.9 ) is positive an sin( 46.9 ) is negative, the correct value for increasing charge is = (e) Now we want the erivative to be negative an sin to be positive. Thus, we take The charge q after N cycles is obtaine by substituting t = NT = N/' into Eq. 31-5: cos cos / RN / / Qe cos N Rt / RNT / q Qe t Qe N Qe N R / cos. We note that the initial charge (setting N = 0 in the above expression) is q 0 = Q cos, where q 0 = 6. is given (with 3 significant figures unerstoo). onsequently, we write the above result as q exp 0 / N q N R. (a) For N = 5, q 6. exp F/1H (b) For N = 10, q 6. exp F/1H (c) For N = 100, q 6. exp F/1H Since ', we ay write T = / as the perio an 1/ as the angular frequency. The tie require for 50 cycles (with 3 significant figures unerstoo) is 3 6 t 50T H F s. The axiu charge on the capacitor ecays accoring to q Qe Rt / ax (this is calle the exponentially ecaying aplitue in Section 31-5), where Q is the charge at tie t = 0 (if we take = 0 in Eq. 31-5). Diviing by Q an taking the natural logarith of both sies, we obtain which leas to ln F HG q Q ax Rt KJ

12 1344 HAPTER 31 3 q 0 10 H ln ax ln R t Q s 6. The assuption state at the en of the proble is equivalent to setting = 0 in Eq Since the axiu energy in the capacitor (each cycle) is given by q / ax, where q ax is the axiu charge (uring a given cycle), then we seek the tie for which qax 1 Q Q qax. Now q ax (referre to as the exponentially ecaying aplitue in Section 31-5) is relate to Q (an the other paraeters of the circuit) by Setting q ax q ax Q /, we solve for t: F Rt / qax Rt Qe ln. H G Q K J F H G F H G K J qax 1 t ln ln ln. R Q KJ R R The ientities ln( 1/ ) ln 1 ln were use to obtain the final for of the result. 7. THNK With the presence of a resistor in the R circuit, oscillation is ape, an the total electroagnetic energy of the syste is no longer conserve, as soe energy is transferre to theral energy in the resistor. EXPRESS et t be a tie at which the capacitor is fully charge in soe cycle an let q ax 1 be the charge on the capacitor then. The energy in the capacitor at that tie is qax 1 U( t) Q e Rt / where q Qe Rt / ax 1 (see the iscussion of the exponentially ecaying aplitue in Section 31-5). One perio later the charge on the fully charge capacitor is q Qe ax R( t T )/

13 1345 where T, an the energy is q Q U t T e ax R( tt )/ ( ). ANAYZE The fractional loss in energy is Rt / R( tt )/ U U( t) U( t T) e e Rt / U U() t e RT / 1 e. Assuing that RT/ is very sall copare to 1 (which woul be the case if the resistance is sall), we expan the exponential (see Appenix E). The first few ters are: e RT / RT R T 1. f we approxiate ', then we can write T as /. As a result, we obtain U RT RT R 11. U EARN The ratio U / U can be rewritten as U U Q where Q / R (not to confuse Q with charge) is calle the quality factor of the oscillating circuit. A high-q circuit has low resistance an hence, low fractional energy loss. 8. (a) We use = /X c = : f 6 Hz)( F)(30.0 V) 0.83 A. (b) = ( Hz)( F)(30.0 V) =.6 A. 9. (a) The current aplitue is given by = V /X, where X = = f. Since the circuit contains only the inuctor an a sinusoial generator, V =. Therefore, V 30.0V A 95.5 A. X 3 f Hz)( H)

14 1346 HAPTER 31 (b) The frequency is now eight ties larger than in part (a), so the inuctive reactance X is eight ties larger an the current is one-eighth as uch. The current is now 30. (a) The current through the resistor is = ( A)/8 = A = 11.9 A. 30.0V A. R 50.0 (b) Regarless of the frequency of the generator, the current is the sae, A. 31. (a) The inuctive reactance for angular frequency is given by X, an the capacitive reactance is given by X = 1/. The two reactances are equal if = 1/, or 1/. The frequency is f H)(10 10 F) Hz. (b) The inuctive reactance is X = = f = (650 Hz)( H) = 4. The capacitive reactance has the sae value at this frequency. (c) The natural frequency for free oscillations is f /, the sae as we foun in part (a). 3. (a) The circuit consists of one generator across one inuctor; therefore, = V. The current aplitue is 5.0 V X (377 ra/s)(1.7 H) A. (b) When the current is at a axiu, its erivative is zero. Thus, Eq gives = 0 at that instant. State another way, since (t) an i(t) have a 90 phase ifference, then (t) ust be zero when i(t) =. The fact that = 90 = / ra is use in part (c). (c) onsier Eq with /. n orer to satisfy this equation, we require sin( t) = 1/. Now we note that the proble states that is increasing in agnitue, which (since it is alreay negative) eans that it is becoing ore negative. Thus, ifferentiating Eq with respect to tie (an eaning the result be negative) we

15 1347 ust also require cos( t) < 0. These conitions iply that t ust equal (n 5/6) [n = integer]. onsequently, Eq yiels (for all values of n) F HG K J F H G KJ i sin n 3 ( A) THNK Our circuit consists of an ac generator that prouces an alternating current, as well as a loa that coul be purely resistive, capacitive, or inuctive. The nature of the loa can be eterine by the phase angle between the current an the ef. EXPRESS The generator ef an the current are given by sin( / 4), i( t) sin( 3 / 4). The expressions show that the ef is axiu when sin( t /4) = 1 or t /4 = (/) ± n [n = integer]. Siilarly, the current is axiu when sin( t 3/4) = 1, or t 3/4 = (/) ± n [n = integer]. ANAYZE (a) The first tie the ef reaches its axiu after t = 0 is when t /4 = / (that is, n = 0). Therefore, A. 3 3 t ra / s) s. (b) The first tie the current reaches its axiu after t = 0 is when t 3/4 = /, as in part (a) with n = 0. Therefore, 5 5 t ra/s) s. (c) The current lags the ef by / ra, so the circuit eleent ust be an inuctor. () The current aplitue is relate to the voltage aplitue V by V = X, where X is the inuctive reactance, given by X =. Furtherore, since there is only one eleent in the circuit, the aplitue of the potential ifference across the eleent ust be the sae as the aplitue of the generator ef: V =. Thus, = an 30.0 V H. 3 (6010 A)(350 ra/s)

16 1348 HAPTER 31 EARN The current in the circuit can be rewritten as 3 i( t) sin sin 4 4 where /. n a purely inuctive circuit, the current lags the voltage by (a) The circuit consists of one generator across one capacitor; therefore, = V. onsequently, the current aplitue is 6 (377 ra / s)( F)(5.0 V) A. X (b) When the current is at a axiu, the charge on the capacitor is changing at its largest rate. This happens not when it is fully charge (±q ax ), but rather as it passes through the (oentary) states of being uncharge (q = 0). Since q = V, then the voltage across the capacitor (an at the generator, by the loop rule) is zero when the current is at a axiu. State ore precisely, the tie-epenent ef (t) an current i(t) have a = 90 phase relation, iplying (t) = 0 when i(t) =. The fact that = 90 = / ra is use in part (c). (c) onsier Eq. 3-8 with 1. n orer to satisfy this equation, we require sin( t) = 1/. Now we note that the proble states that is increasing in agnitue, which (since it is alreay negative) eans that it is becoing ore negative. Thus, ifferentiating Eq. 3-8 with respect to tie (an eaning the result be negative) we ust also require cos( t) < 0. These conitions iply that t ust equal (n 5/6) [n = integer]. onsequently, Eq yiels (for all values of n) or 3 i n i A. 3 sin ( A) A, 35. The resistance of the coil is relate to the reactances an the phase constant by Eq Thus, X X 1/ tan, R R which we solve for R: R ( Hz( H) 6 tan tan 75 ( Hz)( F 89.

17 (a) The circuit has a resistor an a capacitor (but no inuctor). Since the capacitive reactance ecreases with frequency, then the asyptotic value of Z ust be the resistance: R = 500. (b) We escribe three ethos here (each using inforation fro ifferent points on the graph): etho 1: At = 50 ra/s, we have Z 700, which gives = ( Z - R ) 1 = 41 F. etho : At = 50 ra/s, we have X 500, which gives = ( X ) 1 = 40 F. etho 3: At = 50 ra/s, we have X 100, which gives = ( X ) 1 = 40 F. 37. The rs current in the otor is rs 40V rs rs Z R X A. 38. (a) The graph shows that the resonance angular frequency is 5000 ra/s, which eans (using Eq. 31-4) = ( ) 1 = [(5000) ] 1 = 8.0 F. (b) The graph also shows that the current aplitue at resonance is 4.0 A, but at resonance the ipeance Z becoes purely resistive (Z = R) so that we can ivie the ef aplitue by the current aplitue at resonance to fin R: 8.0/4.0 = (a) Now X = 0, while R = 00 an X = 1/f = 177 Therefore, the ipeance is Z R X (00 ) (177 ) 67. (b) The phase angle is 1 X X tan tan 41.5 R 00 (c) The current aplitue is 36.0 V A. Z 67 () We first fin the voltage aplitues across the circuit eleents:

18 1350 HAPTER 31 V V R R (0.135A)(00 ) 7.0V X (0.135A)(177 ) 3.9V The circuit is capacitive, so leas. The phasor iagra is rawn to scale next. 40. A phasor iagra very uch like Fig () leas to the conition: V V = (6.00 V)sin(30º) = 3.00 V. With the agnitue of the capacitor voltage at 5.00 V, this gives a inuctor voltage agnitue equal to 8.00 V. Since the capacitor an inuctor voltage phasors are 180 out of phase, the potential ifference across the inuctor is 8.00 V. 41. THNK We have a series R circuit. Since R,, an are in series, the sae current is riven in all three of the. EXPRESS The capacitive an the inuctive reactances can be written as 1 1 X, X f. f The ipeance of the circuit is by / Z. Z R ( X X ), an the current aplitue is given ANAYZE (a) Substituting the values given, we fin the capacitive reactance to be X f z)( F) Siilarly, the inuctive reactance is X 3 f z)(30 10 H) 86.7.

19 1351 Thus, the ipeance is Z R ( X X ) (00 ) ( ) 06. (b) The phase angle is 1 X X tan tan R 00 (c) The current aplitue is 36.0 V 0.175A. Z 06 () We first fin the voltage aplitues across the circuit eleents: Note that X X V V V R R (0.175 A)(00 ) 35.0 V X (0.175 A)(86.7 ) 15. V X (0.175 A)(37.9 ) 6.6 V, so that leas. The phasor iagra is rawn to scale below. EARN The circuit in this proble is ore inuctive since X X. The phase angle is positive, so the current lags behin the applie ef. 4. (a) Since Z = R + X an X =, then as 0 we fin Z R = 40. (b) = X / = slope = 60 H. 43. (a) Now X = 0, while R = 00 an X = = f = 86.7

20 135 HAPTER 31 both reain unchange. Therefore, the ipeance is Z R X (00 ) (86.7 ) 18. (b) The phase angle is, fro Eq , 1 X X tan tan 3.4. R 00 (c) The current aplitue is now foun to be 36.0 V A. Z 18 () We first fin the voltage aplitues across the circuit eleents: V V R R (0.165 A)(00 ) 33V X (0.165A)(86.7 ) 14.3V. This is an inuctive circuit, so leas. The phasor iagra is rawn to scale next. 44. (a) The capacitive reactance is X f Hz)( F) (b) The ipeance is Z R ( X X ) R ( f X ) 3 (0 ) [ Hz)( H) 16.6 ] 4. (c) The current aplitue is 0V 051 A. Z 4.

21 1353 () Now X 1 eq. Thus, X increases as eq ecreases. (e) Now eq = /, an the new ipeance is Z 3 (0 ) [ Hz)( H) (16.6 )] Therefore, the ipeance ecreases. (f) Since Z 1, it increases. 45. (a) Yes, the voltage aplitue across the inuctor can be uch larger than the aplitue of the generator ef. (b) The aplitue of the voltage across the inuctor in an R series circuit is given by V X. At resonance, the riving angular frequency equals the natural angular frequency: 1/. For the given circuit X 1.0 H (1.0 H)( F) At resonance the capacitive reactance has this sae value, an the ipeance reuces siply: Z = R. onsequently, 10 V 1.0 A. Z R 10 resonance The voltage aplitue across the inuctor is therefore V X 3 (1.0A)(1000 ) V which is uch larger than the aplitue of the generator ef. 46. (a) A sketch of the phasor iagra is shown to the right. (b) We have R = X, or 1 R = X R = which yiels 1 1 f 159 Hz. 5 R (50.0 )( F) (c) = tan 1 (V /V R ) = 45.

22 1354 HAPTER 31 () = 1/R = ra/s. (e) = (1 V)/ R + X = 6/(5 ) 170 A. 47. THNK n a riven R circuit, the current aplitue is axiu at resonance, where the riven angular frequency is equal to the natural angular frequency. EXPRESS For a given aplitue of the generator ef, the current aplitue is given by Z. R ( 1/ ) To explicitly show that is axiu when 1/, we ifferentiate with respect to an set the erivative to zero: 1 1 3/ ( E) [ R ( 1/ ) ]. The only factor that can equal zero is when ( 1 / ), or 1/. ANAYZE (a) For this circuit, the riving angular frequency is ( 100. H)( F) 4 ra / s. (b) When, the ipeance is Z = R, an the current aplitue is 30.0 V 6.00 A. R 5.00 (c) We want to fin the (positive) values of for which / R:. R ( 1/ ) R This ay be rearrange to give F HG 1 KJ 3R.

23 1355 Taking the square root of both sies (acknowleging the two ± roots) an ultiplying by, we obtain ( ) 3 R 1 0. Using the quaratic forula, we fin the sallest positive solution 6 3R 3 R 4 3(0.010 F)(5.00 ) 6 19 ra/s. (1.00 H)( F) 6 6 3( F) (5.00 ) 4(1.00 H)(0.010 F) 6 (1.00 H)( F) () The largest positive solution 6 3R 3 R 4 3(0.010 F)(5.00 ) ra/s. (1.00 H)( F) 6 6 3( F) (5.00 ) 4(1.00 H)(0.010 F) 6 (1.00 H)( F) (e) The fractional with is 1 8 ra/s 19 ra/s ra/s EARN The current aplitue as a function of is plotte below. We see that is a axiu at ra/s an 8 ra/s. 4 ra/s, an is at half axiu (3 A) at (a) With both switches close (which effectively reoves the resistor fro the circuit), the ipeance is just equal to the (net) reactance an is equal to

24 1356 HAPTER 31 X net = (1 V)/(0.447 A) = With switch 1 close but switch open, we have the sae (net) reactance as just iscusse, but now the resistor is part of the circuit; using Eq we fin X net 6.85 R 100. tan tan15 (b) For the first situation escribe in the proble (both switches open) we can reverse our reasoning of part (a) an fin X net first = We observe that the effect of switch 1 iplies R tan= (100 ) tan( 30.9º) = X = X net X net first = 6.85 ( ) = Then Eq leas to = 1/X = 30.6 F. (c) Since X net = X X, then we fin = X / = 301 H. 49. (a) Since eq = 1 + an eq = for the circuit, the resonant frequency is 1 1 eq eq H H F F F 796 Hz (b) The resonant frequency oes not epen on R so it will not change as R increases. (c) Since ( 1 + ) 1/, it will ecrease as 1 increases. () Since 1/ eq an eq ecreases as 3 is reove, will increase. 50. (a) A sketch of the phasor iagra is shown to the right. (b) We have V R = V, which iplies R = X R =

25 1357 which yiels f = / = R/ = 318 Hz. (c) = tan 1 (V /V R ) = +45. () = R/ = ra/s. (e) = (6 V)/ R + X = 3/(40 ) 53.0 A. 51. THNK n a riven R circuit, the current aplitue is axiu at resonance, where the riven angular frequency is equal to the natural angular frequency. t then falls off rapily away fro resonance. EXPRESS We use the expressions foun in Proble 31-47: 3 R 3 R , R R. The resonance angular frequency is 1/. ANAYZE Thus, the fractional half with is R 1 3 R 3. EARN Note that the value of / increases linearly with R; that is, the larger the resistance, the broaer the peak. As an exaple, the ata of Proble gives F H This is in agreeent with the result of Proble The etho use there, however, gives only one significant figure since two nubers close in value are subtracte ( 1 ). Here the subtraction is one algebraically, an three significant figures are obtaine. 5. Since the ipeance of the volteter is large, it will not affect the ipeance of the circuit when connecte in parallel with the circuit. So the reaing will be 100 V in all three cases. 53. THNK Energy is supplie by the 10 V rs ac line to keep the air conitioner running. EXPRESS The ipeance of the circuit is energy elivery is Z R ( X X ), an the average rate of

26 1358 HAPTER 31 rs rsr avg rs. P R R Z ANAYZE (a) Substituting the values given, the ipeance is Z (b) The average rate at which energy has been supplie is 10 V R Z rs avg P Z W W. EARN n a steay-state operation, the total energy store in the capacitor an the inuctor stays constant. Thus, the net energy transfer is fro the generator to the resistor, where electroagnetic energy is issipate in the for of theral energy. 54. The aplitue (peak) value is b g V ax V 100 rs V 141V. 55. The average power issipate in resistance R when the current is alternating is given by P R, where rs is the root-ean-square current. Since /, where is avg rs the current aplitue, this can be written P avg = R/. The power issipate in the sae resistor when the current i is irect is given by P i R. Setting the two powers equal to each other an solving, we obtain. 60A i 184. A. 56. (a) The power consue by the light bulb is P = R/. So we ust let P ax /P in = (/ in ) = 5, or F HG Z Z R / in ax ax inkj F H G / Zax K J F H G Zin K J G R F H b g rs K. J 5 We solve for ax : ax b g b g R 10V / 1000W H Hz (b) Yes, one coul use a variable resistor.

27 1359 (c) Now we ust let or F HG R R R KJ ax bulb 5, bulb i i b g Rax 5 1 R V bulb W () This is not one because the resistors woul consue, rather than teporarily store, electroagnetic energy. 57. We shall use b R R. avg Z R 1/ P where Z R 1/ g is the ipeance. (a) onsiere as a function of, P avg has its largest value when the factor 1/ has the sallest possible value. This occurs for 1/, or R Hz b g b g c Hh F. The circuit is then at resonance. (b) n this case, we want Z to be as large as possible. The ipeance becoes large without boun as becoes very sall. Thus, the sallest average power occurs for = 0 (which is not very ifferent fro a siple open switch). (c) When = 1/, the expression for the average power becoes R avg, so the axiu average power is in the resonant case an is equal to P 30.0V Pavg 90.0 W () At axiu power, the reactances are equal: X = X. The phase angle in this case ay be foun fro

28 1360 HAPTER 31 which iplies = 0. X X tan 0, R (e) At axiu power, the power factor is cos = cos 0 = 1. (f) The iniu average power is P avg = 0 (as it woul be for an open switch). (g) On the other han, at iniu power X 1/ is infinite, which leas us to set tan. n this case, we conclue that = 90. (h) At iniu power, the power factor is cos = cos( 90 ) = This circuit contains no reactances, so rs = rs R total. Using Eq , we fin the average issipate power in resistor R is P R R F HG rs K J r R R. n orer to axiize P R we set the erivative equal to zero: 59. (a) The rs current is rs P r R r R R R r R 0 R r 4 3 R r R r R rs rs Z R f 1/ f Hz5.0H 1/ 550Hz4.70F.59A. 75.0V (b) The rs voltage across R is V R (c) The rs voltage across is V X () The rs voltage across is ab rs.59 A V..59A f 550 Hz 4.70 F rs bc rs 159V.

29 1361 rs rs V X f.59a 550 Hz 5.0H 4 V. c (e) The rs voltage across an together is Vb Vbc Vc 159.5V 3.7 V 64. V. (f) The rs voltage across R,, an together is a ab b V V V 38.8V 64. V 75.0 V. (g) For the resistor R, the power issipate is (h) No energy issipation in. (i) No energy issipation in. P R V 38.8V ab 100 W. R The current in the circuit satisfies i(t) = sin( t ), where an Z R 1/ 45.0 V ra/s9.0 H 1/ 3000 ra/s31. F 1.93A tan 1 tan X X 1/ tan R R ra/s 9.0 H ra/s F (a) The power supplie by the generator is ( ) ( ) sin sin P i t t t t g 1.93A 45.0 V sin 3000 ra/s 0.44 s sin 3000 ra/s 0.44 s W. (b) With

30 136 HAPTER 31 v ( t) V sin( t / ) V cos( t ) c c c where V /, the rate at which the energy in the capacitor changes is c q q Pc i iv t sin t cos t sin t W. c 1.93A sin 3000 ra/s 0.44 s ra/s F (c) The rate at which the energy in the inuctor changes is 1 1 P i i i sin t sin t sin t t t t ra/s 1.93A 9.0H sin 3000ra/s 0.44s W. () The rate at which energy is being issipate by the resistor is PR i R R sin t 1.93A 16.0 sin 3000 ra/s 0.44 s W. (e) Equal. P P P 44.1W 17.0 W 14.4 W 41.5 W P. R c g 61. THNK We have an ac generator connecte to a black box, whose loa is of the for of an R circuit. Given the functional fors of the ef an the current in the circuit, we can euce the nature of the loa. EXPRESS n general, the riving ef an the current can be written as ( t) sin t, i( t) sin( t ). Thus, we have 75 V, = 1.0 A, an 4 for this circuit. The power factor of the circuit is siply given by cos. ANAYZE (a) With = 4.0, we obtain cos = cos( 4.0 ) =

31 1363 (b) Since the phase constant is negative, < 0, t > t. The current leas the ef. (c) The phase constant is relate to the reactance ifference by tan = (X X )/R. We have tan = tan( 4.0 ) = 0.900, a negative nuber. Therefore, X X is negative, which iplies that X > X. The circuit in the box is preoinantly capacitive. () f the circuit were in resonance, X woul be the sae as X, then tan woul be zero, an woul be zero as well. Since is not zero, we conclue the circuit is not in resonance. (e) Since tan is negative an finite, neither the capacitive reactance nor the resistance is zero. This eans the box ust contain a capacitor an a resistor. (f) The inuctive reactance ay be zero, so there nee not be an inuctor. (g) Yes, there is a resistor. (h) The average power is b gb gb g 1 1 Pavg V A W. cos (i) The answers above epen on the frequency only through the phase constant, which is given. f values were given for R,, an, then the value of the frequency woul also be neee to copute the power factor. EARN The phase constant allows us to calculate the power factor an euce the nature of the loa in the circuit. n (f) we state that the inuctance ay be set to zero. f there is an inuctor, then its reactance ust be saller than the capacitive reactance, X < X. 6. We use Eq to fin V s V p F HG N N s p F KJ 500 b H G 100 g K J. 50 V V. 63. THNK The transforer in this proble is a step-own transforer. EXPRESS f N p is the nuber of priary turns, an N s is the nuber of seconary turns, then the step-own voltage in the seconary circuit is

32 1364 HAPTER 31 N s Vs Vp. N p By Oh s law, the current in the seconary circuit is given by V / R. ANAYZE (a) The step-own voltage is V s V p F HG N N (b) The current in the seconary is We fin the priary current fro Eq : p s F HG N N s p s p s F KJ 10 b H G 10 g K J. 500 V 4 V. Vs. 4 V 016. A. R 15 s F KJ 10 b H G 016. g K J. 500 A A. (c) As shown above, the current in the seconary is 0.16A. s s s s EARN n a transforer, the voltages an currents in the seconary circuit are relate to that in the priary circuit by N N s p Vs Vp, s p. N p Ns 64. For step-up transforer: (a) The sallest value of the ratio V / V is achieve by using T T 3 as priary an T 1 T 3 as seconary coil: V 13 /V 3 = ( )/800 = 1.5. s p (b) The secon sallest value of the ratio V / V is achieve by using T 1 T as priary an T T 3 as seconary coil: V 3 /V 13 = 800/00 = s (c) The largest value of the ratio V / V is achieve by using T 1 T as priary an T 1 T 3 as seconary coil: V 13 /V 1 = ( )/00 = s p For the step-own transforer, we siply exchange the priary an seconary coils in each of the three cases above. () The sallest value of the ratio V / V is 1/5.00 = s p p

33 1365 (e) The secon sallest value of the ratio V / V is 1/4.00 = (f) The largest value of the ratio V / V is 1/1.5 = s p s (a) The rs current in the cable is rs P / V t 5010 W / V 315. A. p 3 c h b gb gb g. b315. gbgb0. 60 g 59. Therefore, the rs voltage rop is V rsr 315. A V (b) The rate of energy issipation is P R rs A W. c h 3., so V 3 (c) Now rs W / V 315A b g b g () P 315. A W. 3 3 (e) rs W/ V 31.5 A 31.5A V., so V 31.5A (f) P 31.5A W V. 66. (a) The aplifier is connecte across the priary winings of a transforer an the resistor R is connecte across the seconary winings. (b) f s is the rs current in the seconary coil then the average power elivere to R is P avg s R. Using s N p / Ns p, we obtain N P F p p avg R H G N K J. Next, we fin the current in the priary circuit. This is effectively a circuit consisting of a generator an two resistors in series. One resistance is that of the aplifier (r), an the other is the equivalent resistance R eq of the seconary circuit. Therefore, s p rs rs r Req r N / N R p si where Eq is use for R eq. onsequently, ( N / N ) R p s avg r ( N p / Ns) R P.

34 1366 HAPTER 31 Now, we wish to fin the value of N p /N s such that P avg is a axiu. For brevity, let x = (N p /N s ). Then Rx P avg br xrg, an the erivative with respect to x is This is zero for P avg 3 x b b g g R r xr. r xr x r / R 1000 / We note that for sall x, P avg increases linearly with x, an for large x it ecreases in proportion to 1/x. Thus x = r/r is inee a axiu, not a iniu. Recalling x = (N p /N s ), we conclue that the axiu power is achieve for N / N x 10. p s The iagra that follows is a scheatic of a transforer with a ten to one turns ratio. An actual transforer woul have any ore turns in both the priary an seconary coils. 67. (a) et t / / (b) et t / 4 / 3 4 to obtain t 3 / 4 3 / ra/s s. t / 4 / ra/s.410 s. 3 to obtain (c) Since i leas in phase by /, the eleent ust be a capacitor. () We solve fro X b g 1 / : A 30.0 V350 ra/s F.

35 (a) We observe that = 1566 ra/s. onsequently, X = 754 an X = 199. Hence, Eq gives 1F X X tan K J 1. ra. R (b) We fin the current aplitue fro Eq : HG R ( X X ) 0. 88A. 69. (a) Using = f, X =, X = 1/ an tan() = (X X )/R, we fin = tan 1 [( )/40.0] = ra. (b) Equation gives = 10/ 40 + (16-33) = A. (c) X > X capacitive. 70. (a) We fin fro X f : f 3 X H Hz. (b) The capacitance is foun fro X = () 1 = (f) 1 : 1 1 fx Hz F. (c) Noting that X f an X f 1, we conclue that when f is ouble, X oubles an X reuces by half. Thus, () X = / = X = ( ) = (a) The ipeance is Z = (80.0 V)/(1.5 A) = (b) We can write cos = R/Z. Therefore, R = (64.0 )cos(0.650 ra) = (c) Since the current leas the ef, the circuit is capacitive.

36 1368 HAPTER (a) Fro Eq , we have F HG KJ 1 V V 1 V ( V / 150. ) tan tan V ( V /. 00) which becoes tan 1 (/3 ) = 33.7 or ra. (b) Since > 0, it is inuctive (X > X ). R (c) We have V R = R = 9.98 V, so that V =.00V R = 0.0 V an V = V /1.50 = 13.3 V. Therefore, fro Eq , we have F HG KJ V ( V V ) (9.98 V) (0.0 V 13.3 V) 1.0 V. R 73. (a) Fro Eq. 31-4, we have = ( ) 1 = ((f) ) 1 =.41 H. (b) The total energy is the axiu energy on either evice (see Fig. 31-4). Thus, we have U ax = 1 = 1.4 pj. (c) Of several ethos available to o this part, probably the one ost in the spirit of this proble (consiering the energy that was calculate in part (b)) is to appeal to U ax = 1 Q / (fro hapter 6) to fin the axiu charge: Q = U ax = 8. n. 74. (a) Equation 31-4 irectly gives 1/ ra/s. (b) Equation 16-5 then yiels T = /1.09 s. (c) Although we o not show the graph here, we escribe it: it is a cosine curve with aplitue 00 an perio given in part (b). 75. (a) The ipeance is 15V Z A (b) Fro V R R cos, we get b g b g cos 15V cos ra R A b (c) Since X X sin sin rag, we conclue that X < X. The circuit is preoinantly capacitive. 76. (a) Equation gives f = / = (X ) 1 = 8.84 khz.

37 1369 (b) Because of its inverse relationship with frequency, the reactance will go own by a factor of when f increases by a factor of. The answer is X = THNK The three-phase generator has three ac voltages that are 10 out of phase with each other. EXPRESS To calculate the potential ifference between any two wires, we use the following trigonoetric ientity: where an are any two angles. sin sin sin cos, ANAYZE (a) We consier the following cobinations: V 1 = V 1 V, V 13 = V 1 V 3, an V 3 = V V 3. For V 1, F HG K J F HG K J b 10 t 10 V1 A sin( t) Asin( t 10 ) Asin cos 3A cos t 60 g where sin Siilarly, an V13 Asin( t) Asin ( t 40 ) Asin cos 3Acos t10 40 t t 360 V3 Asin( t 10 ) Asin ( t 40 ) Asin cos 3Acos t180. All three expressions are sinusoial functions of t with angular frequency. (b) We note that each of the above expressions has an aplitue of 3 A. EARN A three-phase generator provies a soother flow of power than a single-phase generator. 78. (a) The effective resistance R eff satisfies R P R eff rs eff echanical, or P hp 746W / hp echanical A rs b gb g b g

38 1370 HAPTER 31 (b) This is not the sae as the resistance R of its coils, but just the effective resistance for power transfer fro electrical to echanical for. n fact rsr woul not give P echanical but rather the rate of energy loss ue to theral issipation. 79. THNK The total energy in the circuit is the su of electrical energy store in the capacitor, an the agnetic energy store in the inuctor. Energy is conserve. EXPRESS et U E be the electrical energy in the capacitor an U B be the agnetic energy in the inuctor. The total energy is U = U E + U B. When U E = 0.500U B (at tie t), then U B =.00U E an U = U E + U B = 3.00U E. Now, U E is given byq /, where q is the charge on the capacitor at tie t. The total energy U is given by Q the axiu charge on the capacitor. /, where Q is ANAYZE (a) Thus, Q 3.00q Q q Q (b) f the capacitor is fully charge at tie t = 0, then the tie-epenent charge on the capacitor is given by q Qcos t. This iplies that the conition q = 0.577Q is satisfie when cost = 0.557, or t = ra. Since / T (where T is the perio of oscillation), t T / T, or t / T = EARN The fraction of total energy that is of electrical nature at a given tie t is given by UE ( Q / )cos t t cos t cos U Q / T. A plot of UE / U as a function of t/ T is given below. Fro the plot, we see that UE / U 1/ 3 at t / T = (a) The reactances are as follows:

39 1371 X X f (400 Hz)(0.04 H) ( f) [ (400 Hz)( F)] 3.88 Z R ( X X ) (0.0 ) ( ) With 90.0 V, we have 90.0 V.6 A.6 A rs 1.85 A. Z Therefore, the rs potential ifference across the resistor is V R rs = rs R = 37.0 V. (b) Across the capacitor, the rs potential ifference is V rs = rs X = 60.9 V. (c) Siilarly, across the inuctor, the rs potential ifference is V rs = rs X = 113 V. () The average rate of energy issipation is P avg = ( rs ) R = 68.6 W. 81. THNK Since the current lags the generator ef, the phase angle is positive an the circuit is ore inuctive than capacitive. EXPRESS et V be the axiu potential ifference across the inuctor, V be the axiu potential ifference across the capacitor, an V R be the axiu potential ifference across the resistor. The phase constant is given by V V VR 1 tan. The axiu ef is relate to the current aplitue by Z, where Z is the ipeance. ANAYZE (a) With V V /.00 an V V /.00, we fin the phase constant to be R V V /.00 V / tan tan (b) The resistance is relate to the ipeance by R Zcos. Thus, 30.0Vcos 45 cos R A EARN With R an known, the inuctive an capacitive reactances are, respectively, X.00R 141, an X R Siilarly, the ipeance of the circuit is

40 137 HAPTER 31 Z 3 (30.0 V) /( A) Fro U ax = 1 we get = A. 83. Fro Eq we get f = 1/ = 1.84 khz. 84. (a) With a phase constant of 45º the (net) reactance ust equal the resistance in the circuit, which eans the circuit ipeance becoes Z = R R = Z/ = 707. (b) Since f = 8000 Hz, then = (8000) ra/s. The net reactance (which, as observe, ust equal the resistance) is therefore X X = ( ) 1 = 707. We are also tol that the resonance frequency is 6000 Hz, which (by Eq. 31-4) eans ( f ) 4 f 4 (6000 Hz). Substituting this for in our previous expression (for the net reactance) we obtain an equation that can be solve for the self-inuctance. Our result is = 3. H. (c) = (((6000)) ) 1 = 1.9 nf. 85. THNK The current an the charge unergo sinusoial oscillations in the circuit. Energy is conserve. EXPRESS The angular frequency oscillation is relate to the capacitance an inuctance by 1/. The electrical energy an agnetic energy in the circuit as a function of tie are given by The axiu value of U E is q Q UE cos ( t) 1 1 Q U B i Q sin ( t ) sin ( t ). Siilarly, the axiu value of U B is also using Q. Q /, which is the total energy in the circuit, U. Q /, which can also be written as /

41 1373 ANAYZE (a) Using the fact that = f, the inuctance is f Hz F H. (b) The total energy ay be calculate fro the inuctor (when the current is at axiu): 1 1 U c. Hhc. Ah J. (c) We solve for Q fro U 1 Q / : c hc h Q U F J EARN Figure 31-4 of the textbook illustrates the oscillations of electrical an agnetic energies. The total energy U U E UB Q / reains constant. When U E is axiu, U B is zero, an vice versa. 86. Fro Eq , we have (0 V /3.00 A) R X X When the switch is open, we have a series R circuit involving just the one capacitor near the upper right corner. Equation leas to 1 R tan tan( 0 ) tan 0. o Now, when the switch is in position 1, the equivalent capacitance in the circuit is. n this case, we have 1 tan 1 tan10.0. R Finally, with the switch in position, the circuit is siply an circuit with current aplitue Z 1 1

42 1374 HAPTER 31 1 where we use the fact that ( ) in siplifying the square root (this fact is evient fro the escription of the first situation, when the switch was open). We solve for, R an fro the three equations above, an the results are as follows: (a) 10V R 165, tan (.00A) tan ( 0.0 ) o tan 1 10 V tan10.0 (b) H, tano (60.0 Hz)(.00 A) tan ( 0.0 ) (c) an.00 A 1 tan / tan ( )(60.0 Hz)(10 V) 1 tan10.0 / tan( 0.0 ) F (a) Eqs an lea to 1 Q (b) We choose the phase constant in Eq to be, so that i 0 = in Eq ). Thus, the energy in the capacitor is U E q Q t (sin ). Differentiating an using the fact that sin cos = sin, we obtain U t E Q sin t. We fin the axiu value occurs whenever sin t 1, which leas (with n = o integer) to 1 n n n 5 4 t s,.4910 s,. The earliest tie is t s. (c) Returning to the above expression for U E / t with the requireent that sin t 1, we obtain

43 1375 F U t Q E HG K J ax i J / s THNK n this proble, we eonstrate that in a riven R circuit, the energies store in the capacitor an the inuctor stay constant; however, energy is transferre fro the riving ef evice to the resistor. EXPRESS The energy store in the capacitor is given by U E q /. Siilarly, the 1 energy store in the inuctor is i The rate of energy supply by the riving ef UB. evice is P i, where isin( ) an sin. The rate with which energy issipates in the resistor is PR i R. ANAYZE (a) Since the charge q is a perioic function of t with perio T, so ust be U E. onsequently, U E will not be change over one coplete cycle. Actually, U E has perio T/, which oes not alter our conclusion. (b) Since the current i is a perioic function of t with perio T, so ust be U B. (c) The energy supplie by the ef evice over one cycle is t T T T U Pt sin( t )sin( t) t [sin t cos cos t sin ]sin( t) t T cos, where we have use T T T sin ( ), sin( )cos( ) 0. 0 t t 0 t t t () Over one cycle, the energy issipate in the resistor is T U P t R t t R T T sin ( ). R 0 R 0 b g b g (e) Since cos VR / R / R, sae. the two quantities are inee the EARN n solving for (c) an (), we coul have use Eqs an 31-71: By oing so, we fin the energy supplie by the generator to be F H G 1 b g K J Pavg T rs rs cos T T cos

44 1376 HAPTER 31 where we substitute rs / an rs /. Siilarly, the energy issipate by the resistor is F H G 1 b g b g K J Pavg,resistor T rsv R T rs rsr T T R. The sae results are obtaine without any integration. 90. Fro Eq. 31-4, we have = ( ) 1 = ((f) ) 1 = 1.59 F. 91. Resonance occurs when the inuctive reactance equals the capacitive reactance. Reactances of a certain type a (in series) just like resistances. Thus, since the resonance values are the sae for both circuits, we have for each circuit: 1 1, 1 1 an aing these equations we fin Since eq 1 an eq ( 1 ), eq = 1 eq resonance in the cobine circuit. 9. When switch S 1 is close an the others are open, the inuctor is essentially out of the circuit an what reains is an R circuit. The tie constant is = R. When switch S is close an the others are open, the capacitor is essentially out of the circuit. n this case, what we have is an R circuit with tie constant = /R. Finally, when switch S 3 is close an the others are open, the resistor is essentially out of the circuit an what reains is an circuit that oscillates with perio T. Substituting = R an = /R, we obtain T. 93. (a) We note that we obtain the axiu value in Eq when we set 1 1 t s 4 f 4( 60) or 4.17 s. The result is sin( sin( 90 ) V. (b) At t = 4.17 s, the current is

45 1377 Oh s law irectly gives i sin ( t ) sin (90 ( 4.3 )) (0.164A) cos(4.3 ) A A. vr ir (0.1495A)(00 ) 9.9V. (c) The capacitor voltage phasor is 90 less than that of the current. Thus, at t = 4.17 s, we obtain v sin(90 ( 4.3 ) 90 ) X X sin(4.3 ) (0.164A)(177 )sin(4.3 ) 11.9V. () The inuctor voltage phasor is 90 ore than that of the current. Therefore, at t = 4.17 s, we fin v sin(90 ( 4.3 ) 90 ) X X sin(4.3 ) (0.164A)(86.7 )sin(4.3 ) 5.85V. (e) Our results for parts (b), (c) an () a to give 36.0 V, the sae as the answer for part (a).