# A charged capacitor is being discharged through a resistor.? A charged capacitor is being discharged through a resistor. At the end of one time constant the charge has been reduced by (1 – 1/e) = 63% of its initial value. At the end of two time constants the charge has been reduced by what percent of its initial value? A) 82% B) 86% C) 100% D) Between 90% and 100% E) need to know more data to answer the question. When a capacitor having capacitance `C`, charged initially at the maximum voltage `U_0` discharges through a resistor `R` the law that gives the variation of capacitor voltage with time is

`U(t) =U_0*e^(-t/tau)`  where `tau = R*C` is the so called time constant.

The current through the resistor is (as Ohm law states)

`I(t) = U(t)/R = (U_0/R)*e^(-t/tau) =I_max*e^(-t/tau)`

To find the charge on the capacitor at time t we integrate the current over the time

`Q(t) =int_0^tI(t)dt =I_max*int_0^te^(-t/tau)dt =(I_max/tau)*(1-e^(-t/tau))=`

`=Q_max*(1-e^(-t/tau))`

For `t =2*tau` we have

`Q(2tau) =Q_max*(1-e^-2) =Q_max*(1-0.1353) =0.8647*Q_max=`

`=86.5%*Q_max`

The correct answer is B) 86%.

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