A capacitor of value 1 μF and a resistor of 5.45 MΩ are connected in series across a 220 V dc supply through a switch. Calculate the time by which the capacitor will be charged to 60 per cent of the supply voltage.
Time constant, \tau=R C=5.45 \times 10^6 \times 1 \times 10^{-6} \mathrm{sec}
= 5 45. seconds
\begin{aligned} & v_c=V\left(1-e^{\frac{-t}{\tau}}\right) \\ & 0.6 \times 230=230\left(1-e^{\frac{-t}{5.45}}\right) \\ & 0.6=1-e^{\frac{-t}{5.45}} \end{aligned}or, e^{-t / 5.45}=1-0.6=0.4
or, e^{-t / 5.45}=\frac{1}{0.4}=2.5
or, \frac{t}{5.45}=\ln 2.5=0.915
or, t=0.915 \times 5.45=4.98 \text { seconds }