Three capacitors each of capacitance $C$ and of breakdown voltage $\mathrm{V}$ are joined in series. The…
Three capacitors each of capacitance $C$ and of breakdown voltage $\mathrm{V}$ are joined in series. The capacitance and breakdown voltage of the combination will be :
$3 \mathrm{C}, 3 \mathrm{~V}$
$\frac{\mathrm{C}}{3}, \frac{\mathrm{V}}{3}$
$3 \mathrm{C}, \frac{\mathrm{V}}{3}$
$\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$
Solution
In series, $\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}}{3}$,
$\mathrm{V}_{\mathrm{eq}}=3 \mathrm{~V}$