An L - C circuit consists of an inductor with an inductance of 55.0 mH and a cap

Question

An L-C circuit consists of an inductor with an inductance of 55.0 mH and a capacitor with a capacitance of 250 F . The initial charge on the capacitor is 6.50 C , and the initial current in the inductor is zero.

(a)What is the maximum voltage across the capacitor?

V =

(b)What is the maximum current in the inductor?

i =

(c)What is the maximum energy stored in the inductor?

U =

(d)When the current in the inductor has half its maximum value, what is the charge on the capacitor?

q =

(e)When the current in the inductor has half its maximum value, what is the energy stored in the inductor?

U =

Explanation / Answer

L = 55 x 10^-3 H

C = 250 x 10^-6 F

Qo = 6.5 uC

w = 1/sqrt(LC) = 1/sqrt(55*10^-3*250*10^-6) = 269.679 rad/s

f = w / 2pi = 269.679 / (2*pi) = 42.92 Hz

(a) maximum voltage across capacitor = Qo / C = 6.5 x 10^-6 / (250 x 10^-6) = 0.026 V

(b) Maximum current in the inductor = w Qo = 269.679* 6.5 * 10^-6 = 0.00175 A = 1.75 mA

(c) maximum energy stored in the inductor = 0.5 L i^2 = 0.5 * 55 * 10^-3 * 0.00175^2 =8.42 x 10^-8 J

(d) U_L = 0.25 U_L_max

So, U_C = 0.75 U_C_max = 0.75 U_L_max = 0.75 * 8.42 * 10^-8 = 0.5 Q^2 / C

q = sqrt(C*2*0.75*8.42*10^-8) = 5.63 uC

charge on the capacitor = q = 5.63 uC

(e) Energy in the inductor = 0.25 * U_L_max = 0.25 * 8.42 * 10^-8 = 2.1 x 10^-8 J

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