Question

SUPE in An 605 oSUniPhys1 315 WA 036 2/20 Submissions Used In an oscillating LC circuit, L-20.0 mH and C-7.40 μ. At time t = 0 the current is 9.30 ma, the charge on the capacitor is 3.90 μ., and the capacitor is charging. (a) What is the total energy in the circuit? (b) What is the maximum charge on the capacitor? (c) What is the maximum current? (d) If the charge on the capacitor is given by q-Q cos(ut + ф), what is the phase angle ф? ? (e) Suppose the data are the same, except that the capacitor is discharging at t - o. What then is Additional Materials eBook O Type here to search

Answer 1

Jiven L 20 mH 20 x 103 H C 7.4 uF-7.4 x10% F q 3.90 uC 3.90 x106 C i-9.30 mA-9.30x 103 A Solution a) Calculate the total energy in the circuit 2C 3.90 x 10-6 2 x 7.4 x 10- F (05200300.9010 0.5 x 20x103 Hx (9.30x103 A)+ 1.893 x 10-6

b) Calculate the maximum charge on the capacitor とmax = V2 x 7.4 x 10-6 x 1.893 x 10-6 5.293 x106 c c) Calculate the maximum curent: 2U 2 x1.893 x10 L 20 x 10-3 H =1.376x10-A

d) Calculate the charge on the capacitor. when t-o q-Ocos(p φ=cos (o - COS 13.90 x 10- c COS 5.293 x10 5.293 x106 C = ±42.530 φ=I-46.90 e) If phase angle is positive then the charge on the capacitor is decreases. Similarly If phase angle is negative then the charge on the capacitor is increases. 1+46.9" ф