In the given circuit, the switch S is closed at time t = 0. The charge Q on the capacitor at any instant t is given by `Q(t) = Q(1 - e^(alphat)).` Find the value of `Q_0` and `alpha` in terms of given parameters as shown in the circuit below.

Explanation

`(Q_0=(CVR_2)/(R_1+R_2),alpha=(R_1+R_2)/(CR_1R_2))`

`Q_0` is the steady charge stored in the capacitor
`Q_0=c`[PD aceoss capacitor in steady state]
`=c["steady state current through" R_2](R_2)`
`=c(v/(R_1+R_2))R_2 therefore Q_0=(CVR_2)/(R_1+R_2)`
`alpha` is `1//tau_0or1/(C R_("net"))`
Here `R_("net")` is equivalent resistance across capicitor after short circuiting the battery .This
`R_("net")=(R_1R_2)/(R_1+R_2)("as R_1 and R_2 in parallel")`
`alpha=1/(c((R_1R_2)/(R_1+R_2)))=(R_1+R_2)/(C R_1R_2)`