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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, 3V
B
$\frac{C}{3}$, $\frac{V}{3}$
C
3C, $\frac{V}{3}$
D
$\frac{C}{3}$, 3V
Detailed Solution
For capacitors in series: $\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$
$C_{eq} = \frac{C}{3}$
In series the applied voltage divides among the capacitors; since they are identical, each gets one-third of the total.
Each capacitor can withstand V, so the combination can withstand V + V + V = 3V before breakdown.
Hence the capacitance is $\frac{C}{3}$ and the breakdown voltage is 3V.
$C_{eq} = \frac{C}{3}$
In series the applied voltage divides among the capacitors; since they are identical, each gets one-third of the total.
Each capacitor can withstand V, so the combination can withstand V + V + V = 3V before breakdown.
Hence the capacitance is $\frac{C}{3}$ and the breakdown voltage is 3V.