Three capacitors each of capacitance $C$ and of breakdown voltage $V$ are joined in series. The capacitance…
Three capacitors each of capacitance $C$ and of breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be
$\frac{\mathrm{C}}{3}, \frac{\mathrm{V}}{3}$
$3 \mathrm{C}, \frac{\mathrm{V}}{3}$
$\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$
$3 \mathrm{C}, 3 \mathrm{~V}$
Solution
In series arrangement charge on each plate of each capacitor has same magnitude.
The potential difference is distributed inversely in the ratio of capacitors ie,
$\mathrm{V}=\mathrm{V}_1+\mathrm{V}_2+\mathrm{V}_3 \ldots$
Here, $V=3 V$
The equivalent capacitance $\mathrm{C}_s$ is given by
$\begin{aligned}
& \frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots \\
& C_s=\frac{\mathrm{C}}{3}
\end{aligned}$
$\begin{aligned}
& \frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots \\
& C_s=\frac{\mathrm{C}}{3}
\end{aligned}$