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# ch31 - Chapter 31 1(a All the energy in the circuit resides...

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1202 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 2 2 C 2.90 10 6 C 2 2 3.60 10 6 F 1.17 10 6 J. (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 2 U L 2 1.168 10 6 J 75 10 3 H 5.58 10 3 A. 2. (a) We recall the fact that the period is the reciprocal of the frequency. It is helpful to refer also to Fig. 31-1. The values of t when plate A will again have maximum positive charge are multiples of the period: t A nT n f n 2.20 10 3 Hz n 4.55 s , where n = 1, 2, 3, 4, . The earliest time is ( n = 1) t A 4.54 s. (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. 31-1). This is when plate A acquires its most negative charge. From that time onward, this situation will repeat once every period. Consequently, t 1 2 T ( n 1) T 1 2 2 n 1 T 2 n 1 2 f 2 n 1 2 2.2 10 3 Hz 2 n 1 2.27 s , where n = 1, 2, 3, 4, . The earliest time is ( n = 1) t 2.27 s. (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. 31-1). Later this will repeat every half- period (compare steps c and g in Fig. 31-1). Therefore,

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1203 t L T 4 ( n 1) T 2 2 n 1 T 4 2 n 1 1.14 s , where n = 1, 2, 3, 4, . The earliest time is ( n = 1) t 1.14 s. 3. (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 6 00 167 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 one-half of a period, or 3.00 s. 4. We find the capacitance from U Q C 1 2 2 : C Q 2 2 U 2.40 10 6 C 2 2 140 10 6 J 2.06 10 8 F. 5. According to U LI Q C 1 2 2 1 2 2 , the current amplitude is I Q LC 3.00 10 6 C 1.10 10 3 H 4.00 10 6 F 4.52 10 2 A. 6. (a) The angular frequency is k m F x m 8.0N 2.0 10 3 m 0.25kg 126rad s . (b) The period is 1/ f and f = /2 . Therefore, T 2 2  rad s 5.0 10 2 s. (c) From = ( LC ) –1/2 , we obtain C 1 2 L 1 126rad s 2 5.0H 1.3 10 5 F.
CHAPTER 31 1204 7. Table 31-1 provides a comparison of energies in the two systems. From the table, we see the following correspondences: 2 2 2 2 1 , , , , 1 1 1 , .

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