p33_017 - this, we calculate the derivative of q with...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
17. (a) The total energy U is the sum of the energies in the inductor and capacitor: U = U E + U B = q 2 2 C + i 2 L 2 = (3 . 80 × 10 6 C) 2 2(7 . 80 × 10 6 F) + (9 . 20 × 10 3 A) 2 (25 . 0 × 10 3 H) 2 =1 . 98 × 10 6 J . (b) We solve U = Q 2 / 2 C for the maximum charge: Q = 2 CU = p 2(7 . 80 × 10 6 F)(1 . 98 × 10 6 J) = 5 . 56 × 10 6 C . (c) From U = I 2 L/ 2, we ±nd the maximum current: I = r 2 U L = r 2(1 . 98 × 10 6 J) 25 . 0 × 10 3 H =1 . 26 × 10 2 A . (d) If q 0 is the charge on the capacitor at time t =0,then q 0 = Q cos φ and φ =cos 1 µ q Q =cos 1 µ 3 . 80 × 10 6 C 5 . 56 × 10 6 C = ± 46 . 9 . For φ =+46 . 9 the charge on the capacitor is decreasing, for φ = 46 . 9 it is increasing. To check
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: this, we calculate the derivative of q with respect to time, evaluated for t = 0. We obtain Q sin , which we wish to be positive. Since sin(+46 . 9 ) is positive and sin( 46 . 9 ) is negative, the correct value for increasing charge is = 46 . 9 . (e) Now we want the derivative to be negative and sin to be positive. Thus, we take = +46 . 9 ....
View Full Document

Ask a homework question - tutors are online