Capacitors of capacities $\mathrm{C}_1, \mathrm{C}_2$ and $\mathrm{C}_3$ are connected in series. If the…
Capacitors of capacities $\mathrm{C}_1, \mathrm{C}_2$ and $\mathrm{C}_3$ are connected in series. If the combination is connected to a supply of ' $\mathrm{V}$ ' volt, then potential difference across capacitor ' $\mathrm{C}_1$ ' is
In series combination, the equivalent capacitance $\mathrm{C}$ is given by
$\begin{aligned} & \frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3} \\ & \therefore \frac{1}{\mathrm{C}}=\frac{\mathrm{C}_2 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1 \mathrm{C}_2 \mathrm{C}_3}\end{aligned}$
$\therefore \mathrm{C}=\frac{\mathrm{C}_1 \mathrm{C}_2 \mathrm{C}_3}{\mathrm{C}_2 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_2}$
Charge $\mathrm{Q}$ stored by the combination is given by
$\mathrm{Q}=\mathrm{CV}=\frac{\mathrm{C}_1 \mathrm{C}_2 \mathrm{C}_3 \mathrm{~V}}{\mathrm{C}_2 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_2}$
Charge on each capacitor is same. Hence potential difference
across $C_1$ is $V_1=\frac{Q}{C_1}=\frac{C_2 C_3 V}{C_2 C_3+C_1 C_3+C_1 C_2}$