Question

A single-loop circuit consists of a 7.2 Ω resistor, 11.9 H inductor, and a 3.3 μF capacitor. Initially, the capacitor has a charge of 6.0 μC and the current is zero. Calculate the charge on the capacitor N complete cycles later for (a) N = 5, (b) N =10, and (c) N = 100.

Answer 1

Answer:

The charge on the capacitor after N cycles can be determined in following way -

- 7.212 RE/2L Giver R Le 11.9 H C = 3.3 MF Initial value of charge on capacitor, %= 6.0 Mc Now, the charge that capacitor has after time sayt, is given by q- gent/2kos (wt+) eqo is the solution of the differential equation of the oscillation in RLC circuit T= 27 = 2a VIC ... (Period of oscillation) considering number of cycles ,N t = NT = 25N - 2ANVIC wt = w x 2aN - 2AN . cos (wt +b) = cos (28N +d) = cos o After N cycles, YN = De Rr/22 caso whern t=0, N-0 & 9-Q (os. However Yo = 6.0LC acoso = 60MC - N = (6.0 pecjen . -Rt/L LP)

Plugging, t = 21NNIC 9 = (6.Quc) e-NARCI case ! A : After 5 cycles, - 560 MC) exp[- (5*** 72 xf23x10-6")] - 6.0 MC *0.94 95 = 5.65 x10-6. – 5-65 pct. Case! B After 10 cycles, 20 - (S.OMC) exp[-(1067x72x / 33x10*')] Gio Mc x0.89 19,0 = 5.33 Mc] 11.9 case: Tina After too cycles, 9100 - (G'OMC ) exp [-(600x* x72x, 5:3x10-5)] - 6.0 MC 80.304 19100 = 1.82 uc] ! Solution : 95 = 5.65MC) 9 10 = 5:33MC 160 = 1.82MC 2Pe)