Question

The capacitor each having capacitance $C=2\mu F$ are connected with a battery of emf $30V$ as shown in the figure. When the switch $S$ is closed. Find
(a) the amount of charge flown through the battery
(b) the heat generated in the circuit
(c) the energy supplied by the battery
(d) the amount of charge flown through the switch $S$

Answer 1

When the switch will close capacitor $2$ and $1$ will be in series and combination will come with parallel to $C_3$

So

$C_{eff}=\frac{1\times 2}{1+2}=\frac{2}{3}\mu F$

$\left(a\right)$ Charge flown through battery $Q=CV=\frac{2}{3}\times 30=20\mu C$

$\left(b\right)$ Hea generated $=\frac{1}{2}{C}_{eff}{V}^2=\frac{1}{2}\times \frac{2}{3}\times (30{)}^2=300\mu J$

$\left(c\right)$ Energy supplied bt the battery= Heat generated+Work done

$=\frac{1}{2}C{V}^2+\frac{Q^2}{2C}$

$\frac{1}{2}\times \frac{2}{3}(30{)}^2+\frac{(20{)}^2\times 3}{2\times 2}$

$=600\mu J$

$\left(d\right)$ Amount of charges flow through the switch

$=CV=2\times 30$

$=60\mu C$