Question

The time constant of an RC circuit is 0.6 s. a) To what voltage can the capacitor be charged in 1 s if the voltage source is 100 V? b) In what time will the capacitor be discharged from that value to its half?

Answer 1

(a) The voltage:

$$V_C=V_S(1-e^{-\frac{t}{\tau}})=V_S(1-e^{-\frac{1}{0.6}})=0.81V_S,$$

or 81% of the source voltage.

(b) Apply the same equation:

$$V_C=0.5V_S=V_S(1-e^{-\frac{t}{\tau}}),\\\space\\ 0.5=1-e^{-\frac{t}{\tau}},\\\space\\ t=-\tau\ln(1-0.5)=-0.6\ln0.5=0.42\text{ s}.$$