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$3C,$$\frac{V}{3}$
B$\;\frac{C}{3}$$ ,3V$
C$3C,3V$
D$\;\frac{C}{3},\frac{V}{3}$
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B$\;\frac{C}{3}$$ ,3V$
b Three capacitors of capacitance $C$ each are in series
$\therefore$ Total capacitance, $C_{\text {total }}=C / 3$
The charge is the same, $Q,$ when capacitors are in series.
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