A condenser of capacity 5 $\mathrm{\mu }$F is connected to a constant source of e.m.f. 200 volt as shown in fig. What will be the amount of heat produced in R₁, when the key is
thrown from contact 1 to 2 ? 

Consider the position of key in contact 1.
Energy stored in condenser $=\frac{1}{2}{\mathrm{CE}}^2$
Now the key is thrown to contact 2. If, be the current in the circuit, then
$\begin{array}{r}{\mathrm{H}}_1={\mathrm{i}}^2{\mathrm{R}}_1\text{ and }{\mathrm{H}}_2={\mathrm{i}}^2{\mathrm{R}}_2\\ \therefore \mathrm{H}={\mathrm{H}}_1+{\mathrm{H}}_2={\mathrm{i}}^2\left({\mathrm{R}}_1+{\mathrm{R}}_2\right)=\frac{1}{2}{\mathrm{CE}}^2\end{array}$
or ${\mathrm{i}}^2(500+330)=\frac{1}{2}\left(5\times {10}^{-6}\right)\times (200{)}^2$
$\begin{array}{r}{\mathrm{i}}^2=\frac{\left(6\times {10}^{-6}\right)\left(100\right)\left(200\right)}{830}\\ \therefore {\mathrm{H}}_1=\frac{\left(5\times {10}^{-6}\right)\left(100\right)\left(200\right)\times 500}{830\times } \mathrm{J}=0.06\mathrm{J}\end{array}$