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
Charge on each capacitor is same. Hence potential difference
across \(C_1\) is \(V_1=\frac{Q}{C_1}=\frac{C_2 C_3 V}{C_2 C_3+C_1 C_3+C_1 C_2}\)