An L - C circuit consists of an inductor with an inductance of 65.0 mH and a cap

Question

An L-C circuit consists of an inductor with an inductance of 65.0 mH and a capacitor with a capacitance of 200 F . The initial charge on the capacitor is 6.00 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

(A) Q = C V

6 uC = (200 uF ) (V)

V = 0.03 Volt

(B) C V^2 / 2 = L I^2 / 2

(200 x 10^-6) (0.03^2) = (0.065) (I^2)

I = 1.66 x 10^-3 A or 1.66 mA

(C) U = (200 x 10^-6)(0.03^2) /2

= 90 x 10^-9 J

(d) Q = sqrt(3) Q0 / 2 = 1.732 x 6 / 2

Q =5.20 uC

(e) L = U0/4 = 22.5 x 10^-9 J

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