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?

Explanation:

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

representative si the final answer! : )

explanation: in any substance/object, the particles inside it (atoms/molecules) constantly move in random directions and with random speeds (this motion is called brownian motion). as a result, the particles have some kinetic energy (which is proportional to the temperature of the substance).

explanation:

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