sinb2ωtg to write this as

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Aubrey Fitzgerald
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1 Chapter (a) The charge (as a function of tie) is given by q= Qsinωt, where Q is the axiu charge on the capacitor an ω 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 = ωq. Since ω = 1/ LC, q i = =ωqcos ωt, t b g c hc h Q= LC =. 00A H F = C. (b) The energy store in the capacitor is given by q Q t U = E C = sin ω C an its rate of change is U Q E ω sinωtcosωt = t C We use the trigonoetric ientity cosωtsin ωt = 1 sinbωtg to write this as U E Q = ω sin ωt. t C The greatest rate of change occurs when sin(ωt) = 1 or ωt = π/ ra. This eans π π π t= = LC = ( )( ) = 4ω 4 4 b g H F s. (c) Substituting ω = π/t an sin(ωt) = 1 into U E /t = (ωq /C) sin(ωt), we obtain U E πq πq = =. t TC TC ax c hc h so Now T = π LC = π H F = s,

2 U E t ax 4 π ( C) 4 6 ( s)( 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/ (a) The axiu charge is Q = CV ax = ( F)(3.0 V) = C. (b) Fro U = 1 1 L = Q / C we get Q = = LC C 3 9 c Hhc Fh = A. (c) When the current is at a axiu, the agnetic energy is at a axiu also: c hc h UB,ax = L = H A = J. 33. (a) The generator ef an the current are given by = sin( ω π / 4), it ( ) = sin( ω 3 π / 4). The expressions show that the ef is axiu when sin(ω t π/4) = 1 or ω t π/4 = (π/) ± nπ [n = integer]. The first tie this occurs after t = 0 is when ω t π/4 = π/ (that is, n = 0). Therefore, 3π 3π t = = = 4ω 4(350 ra/s) s. (b) The current is axiu when sin(ω t 3π/4) = 1, or ω t 3π/4 = (π/) ± nπ [n = integer]. The first tie this occurs after t = 0 is when ω t 3π/4 = π/ (as in part (a), n = 0). Therefore, 5π 5π t = = = s. 4ω 4(350 ra/s) (c) The current lags the ef by + π / ra, so the circuit eleent ust be an inuctor.

3 () The current aplitue is relate to the voltage aplitue V L by V L = X L, where X L is the inuctive reactance, given by X L = ω L. 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 L =. Thus, = ω L an 30. 0V L = = 3 ω (60 10 A)(350 ra / s) = H. Note: The current in the circuit can be rewritten as 3π π it () = sin ω = sin ω φ 4 4 where φ =+ π /. n a purely inuctive circuit, the current lags the voltage by The resistance of the coil is relate to the reactances an the phase constant by Eq Thus, which we solve for R: X X R ω L 1/ ω C = R L C = tan φ, R= ω L = ( Hz( H) 6 tanφ ωc tan 75 π)(930 ( Hz)( F π)(930 = 89 Ω. 44. (a) The capacitive reactance is X C 1 1 = = = 16.6 Ω. 6 π fc π(400 Hz)( F) (b) The ipeance is Z = R + ( X X ) = R + ( π fl X ) L C C = Ω + (400 Ω = Ω 3 (0 ) [π Hz)( H) 16.6 ] 4. (c) The current aplitue is

4 () Now X C C 0V = = = 051. A. Z 4Ω 1 eq. Thus, X C increases as C eq ecreases. (e) Now C eq = C/, an the new ipeance is Z = Ω + π(400 Ω = Ω< Ω 3 (0 ) [ Hz)( H) (16.6 )] Therefore, the ipeance ecreases. (f) Since Z 1, it increases. 57. We shall use b R R = =. avg Z R + ( ωl 1/ ωc) where Z = R + ωl 1/ ωcg is the ipeance. P R + ω L 1/ ω C (a) Consiere as a function of C, P avg has its largest value when the factor ( ) has the sallest possible value. This occurs for ω L= 1/ ω C, or 1 1 C = = ω L π 600. Hz b gb gc Hh 3 = 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 C becoes very sall. Thus, the sallest average power occurs for C = 0 (which is not very ifferent fro a siple open switch). (c) When ω L = 1/ω C, the expression for the average power becoes R avg =, so the axiu average power is in the resonant case an is equal to P

5 avg ( 30.0V) 5.00 ( Ω) P = = 90.0 W. () At axiu power, the reactances are equal: X L = X C. The phase angle φ in this case ay be foun fro XL XC tan φ = = 0, R which iplies φ = 0. (e) At axiu power, the power factor is cos φ = 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 C 1/C is infinite, which leas us to set tanφ =. n this case, we conclue that φ = 90. (h) At iniu power, the power factor is cos φ = cos( 90 ) = The current in the circuit satisfies i(t) = sin(ω t φ), where = = = Z R + L C ( ω 1/ ω ) 45.0 V { } ( 16.0 ) ( 3000 ra/s)( 9.0 H) 1/ ( 3000ra/s)( 31. μf) = 1.93A Ω + an 1 XL XC 1 ωl 1/ ωc φ = tan = tan R R = ( )( ) 3000ra/s 9.0H Ω ( 3000 ra/s)( 16.0 Ω)( 31. μf) 1 = tan (a) The power supplie by the generator is () () sin( ω φ) sinω ( )( ) ( )( ) ( )( ) 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 46.5 = 41.4 W.

6 (b) With v ( t) = V sin( ω t φ π / ) = V cos( ω t φ) c c c where V = / ω C, the rate at which the energy in the capacitor changes is c q q Pc = = i = ivc t C C sin ( ω t φ = ) cos( t ) sin ( t ) ωc ω φ = ωc ω φ ( 1.93A) 6 ( )( ) = 17.0 W. ( )( ) ( ) = sin 3000 ra/s 0.44 s ra/s F (c) The rate at which the energy in the inuctor changes is 1 1 PL = Li = Li i = Lsin ( ωt φ) sin ( t ) L sin ( t ) t t t ω φ = ω ω φ 1 = ( 3000ra/s )( 1.93A ) ( 9.0H ) sin ( 3000ra/s )( 0.44s ) ( 46.5 ) = 44.1 W. () The rate at which energy is being issipate by the resistor is ( ω φ) ( ) ( ) ( )( ) PR = i R= Rsin t = 1.93A 16.0 Ω sin 3000 ra/s 0.44 s 46.5 = 14.4 W. (e) Equal. P + P + P = 44.1W 17.0 W W = 41.5 W = P. L R c g

Physics 4B. Chapter 31: Questions: 2, 8, 12 Exercises & Problems: 2, 23, 24, 32, 41, 44, 48, 60, 72, 83. n n f

Physics 4B. Chapter 31: Questions: 2, 8, 12 Exercises & Problems: 2, 23, 24, 32, 41, 44, 48, 60, 72, 83. n n f Physics 4B Solutions to hapter 1 HW hapter 1: Questions:, 8, 1 Exercises & Probles:,, 4,, 41, 44, 48, 60, 7, 8 Question 1- (a) less; (b) greater Question 1-8 (a) 1 an 4; (b) an Question 1-1 (a) lea; (b)