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

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1 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) capacitive; (c) less Proble 1- (a) We recall the fact that the perio is the reciprocal of the frequency. t is helpful to refer also to ig The values of t when plate A will again have axiu positive charge are ultiples of the perio: n n ta = nt = = f b = n 500. μs, Hz where n = 1,,, 4,. The earliest tie is (n = 1) t = 5.00μs. (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 ig. 1-1). This is when plate A acquires its ost negative charge. ro that tie onwar, this situation will repeat once every perio. onsequently, ( n1) ( n1) 1 1 t= T + ( n 1) T = ( n 1) T = = = ( n1)(.50μs), f 10 Hz A ( ) where n = 1,,, 4,. The earliest tie is (n = 1) t =.50μs. (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 ig. 1-1). ater this will repeat every half-perio (copare steps c an g in ig. 1-1). Therefore, T ( n1) T T t = + = ( n 1) = ( n 1)( 1.5μs ), 4 4 g

2 where n = 1,,, 4,. The earliest tie is (n = 1) t = 1.5μs. Proble 1- (a) The total energy U is the su of the energies in the inuctor an capacitor: ( ) A H J. q i U = UE + UB = + = + = (b) We solve U = Q / for the axiu charge: c hc h. Q= U = J = (c) ro U = /, we fin the axiu current: c Jh U = = H = A. () f q 0 is the charge on the capacitor at tie t = 0, then q 0 = Q cos φ an φ = cos H G q Q = KJ cos 1 1 HG =± KJ or φ = the charge on the capacitor is ecreasing, for φ = 46.9 it is increasing. To check this, we calculate the erivative of q with respect to tie, evaluate for t = 0. We obtain ωq sin φ, 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 φ = (e) Now we want the erivative to be negative an sin φ to be positive. Thus, we take φ = Proble 1-4 The charge q after N cycles is obtaine by substituting t = NT = πn/ω' into Eq. 1-5: ( ω φ) ω ( π ) Rt / RNT / q = Qe cos t+ = Qe cos N / ω + φ RN ( π / )/ = Qe cos πn + φ = Qe Nπ R / cos φ.

3 We note that the initial charge (setting N = 0 in the above expression) is q 0 = Q cos φ, where q 0 = 6. μ is given (with significant figures unerstoo). onsequently, we write the above result as q exp 0 ( / N q Nπ R ) =. (a) or N = 5, q5 = 6.μ exp 5π 7.Ω /1H = 5.85μ. (b) or N = 10, q10 = 6.μ exp 10π 7.Ω /1H = 5.5μ. (c) or N = 100, q100 = 6.μ exp 100π 7.Ω /1H = 1.9μ. Proble 1- (a) The circuit consists of one generator across one inuctor; therefore, =. The current aplitue is 5.0 = = = = X ω (77 ra/s)(1.7 H) A. (b) When the current is at a axiu, its erivative is zero. Thus, Eq. 0-5 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 φ = 90 = π/ ra is use in part (c). (c) onsier Eq. 1-8 with = /. n orer to satisfy this equation, we require sin(ω 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. 1-8 with respect to tie (an eaning the result be negative) we ust also require cos(ω t) < 0. These conitions iply that ωt ust equal (nπ 5π/6) [n = integer]. onsequently, Eq. 1-9 yiels (for all values of n) HG K J = H G i = sin nπ 5π π ( A) KJ = A. Proble 1-41 (a) The capacitive reactance is X = = = = 7.9 Ω. ω π f π(60.0 Η z)( ) The inuctive reactance 86.7 Ω is unchange. The new ipeance is Z = R + ( X X ) = (00 Ω ) + (7.9Ω86.7 Ω ) = 06 Ω.

4 (b) The phase angle is 1 X X Ω7.9Ω φ = tan = tan = 1.7. R 00Ω (c) The current aplitue is 6.0 = = = 0.175A. Z 06Ω () We first fin the voltage aplitues across the circuit eleents: R = R= (0.175 A)(00 Ω ) = 5.0 = X = (0.175 A)(86.7 Ω ) = 15. = X = (0.175 A)(7.9 Ω ) = 6.6 Note that X > X, so that leas. The phasor iagra is rawn to scale below. Proble 1-44 (a) The capacitive reactance is X 1 1 = = = 16.6 Ω. π f π(400 Hz)( ) (b) The ipeance is Z = R + ( X X ) = R + ( π fx ) = Ω + (400 Ω = Ω (0 ) [π Hz)( H) 16.6 ] 4. (c) The current aplitue is 0 = = = 051. A. Z 4Ω

5 () Now X 1 eq. Thus, X increases as eq ecreases. (e) Now eq = /, an the new ipeance is Z = Ω + π(400 Ω = Ω< Ω (0 ) [ Hz)( H) (16.6 )] Therefore, the ipeance ecreases. (f) Since Z 1, it increases. Proble 1-48 (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 X net = (1 )/(0.447 A) = 6.85 Ω. 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 Ω R = = = 100 Ω. tanφ tan15 (b) or 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 tanφ = (100 Ω) tan( 0.9º) = Ω. X = X net X net first = 6.85 Ω ( Ω) = Ω. Then Eq. 1-9 leas to = 1/ωX = 0.6 μ. (c) Since X net = X X, then we fin = X /ω = 01 H.

6 Proble 1-60 The current in the circuit satisfies i(t) = sin(ω t φ), where = = = Z R + ( ω 1/ ω ) 45.0 { μ ) } ( = 1.9A 16.0 Ω ra/s 9.0 H 1/ 000 ra/s 1. an 1 X X 1 ω1/ ω φ = tan = tan R R = ra/s 9.0 H Ω ( 000 ra/s)( 16.0 Ω)( 1. μ ) 1 = tan (a) The power supplie by the generator is () () sin( ω φ) sinω P = i t t = t t g = 1.9A 45.0 sin 000 ra/s 0.44 s sin 000 ra/s 0.44 s 46.5 = 41.4 W. (b) With v () t = sin( ω tφ π /) = cos( ω t φ) c c c where = / ω, the rate at which the energy in the capacitor changes is c q q Pc = = i = iv t sin ( ω t φ = ) cos( t ) sin ( t ) ω ω φ = ω ω φ ( 1.9A) = sin ( 000 ra/s)( 0.44 s) ( 46.5 ) ( ) 000 ra/s c =17.0 W. (c) The rate at which the energy in the inuctor changes is

7 1 1 P = i = i i = sin ( ωtφ) sin ( t ) sin ( t ) t t t ω φ = ω ω φ 1 = ( 000ra/s )( 1.9A ) ( 9.0H ) sin ( 000ra/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 = 1.9A 16.0 Ω sin 000 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. R c g Proble 1-7 (a) ro Eq. 1-65, we have φ = tan HG = KJ which becoes tan 1 (/ ) =.7 or ra. (b) Since φ > 0, it is inuctive (X > X ). HG 1 1 ( / 150. ) tan ( / 00. ) R (c) We have R = R = 9.98, so that =.00 R = 0.0 an = /1.50 = 1.. Therefore, fro Eq. 1-60, we have KJ = + ( ) = (9.98 ) + ( ) = 1.0. R Proble 1-8 ro Eq. 1-4 we get f = 1/π = 1.84 khz.