A capacitor with capacitance 5 times 10^-5 F is charged by connecting it to a 10

Question

A capacitor with capacitance 5 times 10^-5 F is charged by connecting it to a 10 V battery. The capacitor is disconnected from the battery and connected across an inductor with 2 H inductance. (a) What is the period of the electrical oscillations? (b) How much energy is initially stored in the capacitor? (c) What is the charge on the capacitor 0.02 s after the connection to the inductor is made? (d) At the time given in part (c), what is the current in the inductor? (e) At the time given in part (c), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

Explanation / Answer

Here ,

C = 5 *10^-5 F

V = 10 V

L = 2 H

a) period of oscillation = 2pi * sqrt(L* C)

period of oscillation = 2pi * sqrt(5 *10^-5 * 2)

period of oscillation = 0.063 s

b)

energy stored in the capacitor = 0.50 * C * V^2

energy stored in the capacitor = 0.50 * 5 *10^-5 * 10^2

energy stored in the capacitor = 2.5 *10^-3 J

c)

as charge is given as

Q = 5 *10^-5 * 10 * cos(2pi * t/T)

Q = 5 *10^-5 * 10 * cos(2pi * 0.02/0.063)

Q = 2 *10^-4 C

d)

let the current is I

0.50 * 2 * I^2 + 0.50 * (2 *10^-3)^2/(5 *10^-5) = 2.5 *10^-3

solving for I

I = 0.0458 A

the current is 0.0458 A

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