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 will be
correctly given by 

The given problem may be approached in a simplified manner by the following considerations. Since the resistances R and R are in series 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 $\mathrm{\tau }=\mathrm{RC}/2$ . Thus the growth of potential in the capacitor will be given by
$\mathrm{V}\left(\mathrm{t}\right)=\frac{\mathrm{E}}{2}\left[1-\exp \left(-\frac{\mathrm{t}}{\mathrm{\tau }}\right)\right]=\frac{\mathrm{E}}{2}\left[1-\exp \left(-\frac{2\mathrm{t}}{\mathrm{RC}}\right)\right]$