The capacitor in an RC circuit is discharged with a time constant of 3.00ms. A)

Question

The capacitor in an RC circuit is discharged with a time constant of 3.00ms. A) At what time after the discharge begins is the charge on the capacitor reduced to half its initial value? B) At what time after the discharge begins is the energy stored in the capacitor reduced to half its initial value? The capacitor in an RC circuit is discharged with a time constant of 3.00ms. A) At what time after the discharge begins is the charge on the capacitor reduced to half its initial value? B) At what time after the discharge begins is the energy stored in the capacitor reduced to half its initial value?

Explanation / Answer

a)

Q = Qo*e^(-t/CR)
Qt/Qo = 0.5 =e^(-t/CR)
e^(-t/CR)=0.5
e^(-t/3)=0.5

t=2.07944154168m s

b)

Energy stored on capacitor U = (1/2) * C * V2

Given U = U0 / 2

=> 0.5 * C * V02 / 2 = 0.5 * C * V2

Hence final voltage across the capacitor V = V0 / ?2

During discharge, the voltage across the capacitor

V = V0 * e-t/?

V0/?2 = V0 * e-t/3 ms

et/3 ms = ?2

t / 3 ms = ln ?2

hence t = 3 * ln ?2

= 1.03972077084ms

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