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Unformatted text preview: Chapter 31 1. (a) All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then zero. If Q is the maximum charge on the capacitor, then the total energy is U Q C = = =  2 6 2 6 6 2 2 90 10 2 360 10 117 10 . . . C F J. c h c h (b) When the capacitor is fully discharged, the current is a maximum and all the energy resides in the inductor. If I is the maximum current, then U = LI 2 /2 leads to I U L = = =  2 2 1168 10 75 10 558 10 6 3 3 . . J H A. c h 2. According to U LI Q C = = 1 2 2 1 2 2 , the current amplitude is I Q LC = = =  300 10 4 00 10 4 52 10 6 3 6 2 . . . C 1.10 10 H F A. c hc h 3. We find the capacitance from U Q C = 1 2 2 : C Q U = = =  2 6 2 6 9 2 160 10 2 140 10 914 10 . . C J F. c h c h 4. (a) The period is T = 4(1.50 s) = 6.00 s. (b) The frequency is the reciprocal of the period: f T = = = 1 1 600 167 10 5 . . s Hz. (c) The magnetic energy does not depend on the direction of the current (since U B i 2 ), so this will occur after onehalf of a period, or 3.00 s. 1241 CHAPTER 31 5. (a) We recall the fact that the period is the reciprocal of the frequency. It is helpful to refer also to Fig. 311. The values of t when plate A will again have maximum positive charge are multiples of the period: t nT n f n n A = = = = 2 00 10 500 3 . . , Hz s b g where n = 1, 2, 3, 4, K . The earliest time is ( n =1) 5.00 s. A t = (b) We note that it takes t T = 1 2 for the charge on the other plate to reach its maximum positive value for the first time (compare steps a and e in Fig. 311). This is when plate A acquires its most negative charge. From that time onward, this situation will repeat once every period. Consequently, ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 3 2 1 2 1 1 1 ( 1) 2 1 2 1 2.50 s , 2 2 2 2 2 10 Hz n n t T n T n T n f  = + = = = = where n = 1, 2, 3, 4, K . The earliest time is ( n =1) 2.50 s. t = (c) At t T = 1 4 , the current and the magnetic field in the inductor reach maximum values for the first time (compare steps a and c in Fig. 311). Later this will repeat every half period (compare steps c and g in Fig. 311). Therefore, ( 29 ( 29 ( 29 ( 1) 2 1 2 1 1.25 s , 4 2 4 L T n T T t n n  = + = = where n = 1, 2, 3, 4, K . The earliest time is ( n =1) 1.25 s. t = 6. (a) The angular frequency is = = = = k m F x m 8 0 050 89 13 . . . N 2.0 10 m kg rad s c hb g (b) The period is 1/ f and f = /2 . Therefore, T = = =  2 2 7 0 10 2 89 rad s s. . (c) From = ( LC ) 1/2 , we obtain 218 C L = = =  1 1 89 50 2 5 10 2 2 5 rad s H F. b gb g . . 7. (a) The mass m corresponds to the inductance, so m = 1.25 kg....
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 Fall '08
 Many
 Physics, Charge, Current, Energy

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