In the circuit shown in fig switch S is closed at time t =0.

a. What is the current `I_(0)` leacing the battery at t=0, immediately after the switch is closed?
b. What is the current I "long time" later?
c. What charge has accmulated on the capacitor after this long time?
d. If, finally, swtich S is opened again, how long will it take after the switch is opened for the capactior to lose 80% of its charge?

Explanation

At t =0, capacitor acts as short - circuit. There will not be any
current in the `40Omega` resistance. Therefore,
`I_0 = (10)/(20) = 0.5 A`
b. Capacitor acts as open cirucit, so
`I = (10)/(60) = (10)/(6)A`
c. Voltage across capacitor is
`V_(40) = IR = (1)/(6) xx 40,Q = CV = 0.50 xx(1)/(6)xx40 = (10)/(3)muC`
d. `q = q_0(e ^(-t//RC)),` capacitor will discharge through the `40Omega`
resistor when S is open, i.e.,
`(20)/(100)q_0 = q_0 (e^((t)/(40xx0.5xx10^(-6))))` or `(1)/(5)=(1)/((t)/(e^(20xx10^(-6))))`
or `t=20xx10^(-6)In(5)`