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

• A

  $\frac{\mathrm{C}}{3}, \frac{\mathrm{V}}{3}$
• B

  $3 \mathrm{C}, \frac{\mathrm{V}}{3}$
• Correct answer
  C

  $\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$
• D

  $3 \mathrm{C}, 3 \mathrm{~V}$

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
$$ \begin{aligned} & \frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots \\ & C_s=\frac{\mathrm{C}}{3} \end{aligned} $$

Hence, the correct answer is option: 'C'.