question_answer

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    [AIPMT (S) 2009]

A) \[\frac{C}{3},\,\frac{V}{3}\]

B) \[3C,\,\,\frac{V}{3}\]

C) \[\frac{C}{3},\,3V\]

D) \[3C,\,3V\]

Correct Answer: C

Solution :

[c] In series arrangement charge on each plate of each capacitor has same magnitude.

The potential difference is distributed inversely in the ratio of capacitors i.e.,

\[V={{V}_{1}}+{{V}_{2}}+{{V}_{3}}...\]

Here,     \[V=3V\]

The equivalent capacitance \[{{C}_{s}}\] is given by

\[\frac{1}{{{C}_{s}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+...\]\[{{C}_{s}}=\frac{C}{3}\]