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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 one-half 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. 31-1. 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. 31-1). 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. 31-1). Later this will repeat every half- period (compare steps c and g in Fig. 31-1). 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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 42

CH-31 - Chapter 31 1(a All the energy in the circuit...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online