A capacitor of capacitance 4 µF is discharging through a 2.0-MΩ resistor. At what time will the energy stored in the capacitor be one-third of its initial value?

Question

Kristin Tucker

Physics

27 May, 20:55

A capacitor of capacitance 4 µF is discharging through a 2.0-MΩ resistor. At what time will the energy stored in the capacitor be one-third of its initial value?

+2

Answers (1)

Know the Answer?

Amani Byrd · Answer 1

Given that,

The capacitance of the capacitor is

C = 4 µF

Discharging through a resistor of resistance

R = 2 MΩ

Time the energy stored in the capacitor be one-third of its initial energy

i. e. u (E) _ final = ⅓u (E) _initial

U / Uo = ⅓

Energy stored in a capacitor (discharging) can be determined using

U = Uo•RC•exp (-t/RC)

U / Uo = - RC exp (-t/RC)

U / Uo = ⅓

RC = 2 * 10^6 * 4 * 10^-6 = 8s

⅓ = 8 exp (-t / 8)

Divide both side by - 8

1/24 = exp (-t/8)

Take In of both sides

In (1/24) = In•exp (-t/8)

-3.1781 = - t / 8

t = - 3.1781 * - 8

t = 25.42 seconds

It will take 25.42 seconds for he capacitor to be ⅓ of it's initial energy