In the circuit shown in the diagram, `E` is the e.m.f of the cell, connected to two resistances. Each of magnitude `R` and a capacitor of capacitance `C` as shown in the diagram. If the switch key `K` is closed at time `t=0`, the growth of potential `V` across the capacitor with be correctly given by
.

Options

A. `V(t)=E[1-exp(-(t)/(RC))]`
B. `V(t)=(E )/(2)[1-exp(-(2t)/(RC))]`
C. `V(t)=E[1-exp(-(2t)/(RC))]`
D. `V(t)=(E)/(2)[1-exp(-(t)/(RC))]`

Answer

B

Explanation

The given problem may be approached in a simplified manner of the following considerations. Since the resistances `R` and `R` are in seriea with the battery, the potential difference across `R`(which charges the capacitor) will have maximum value `E//2`. Also, if the battery is absent, and the (charged) capacitor were to discharge, the two resistors `R` and `R` will be connected in parallel (i.e) effective resistance across the capacitor is `(R//2)` and hence the time-constant in the cirstant in the circuit is `tau = RC//2`. Thus the growth of potential in the capacitor will be given by
`V(t)=(E)/(2)[1-exp(-(t)/(tau))]`
`=(E)/(2)[1-exp(-(2t)/(RC))]`.