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?
D) Between 90% and 100%
E) need to know more data to answer the question.
1 Answer | Add Yours
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))=`
For `t =2*tau` we have
`Q(2tau) =Q_max*(1-e^-2) =Q_max*(1-0.1353) =0.8647*Q_max=`
The correct answer is B) 86%.
We’ve answered 319,194 questions. We can answer yours, too.Ask a question