Question:

Three capacitors each of capacitance ' C ' and breakdown voltage ' $V$ ' are connected in series. The capacitance and breakdown voltage of the series combination will be respectively

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For \(n\) identical capacitors in series: \[ C_{\text{eq}}=\frac{C}{n} \] and total breakdown voltage becomes \(nV\).
Updated On: May 14, 2026
  • $3C, 3V$
  • $\frac{C}{3}, \frac{V}{3}$
  • $3C, \frac{V}{3}$
  • $\frac{C}{3}, 3V$
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The Correct Option is D

Solution and Explanation

Concept:
For three identical capacitors in series: \[ \frac{1}{C_{\text{eq}}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C} \] Also, in series combination, voltage divides equally for identical capacitors, so total safe voltage becomes the sum of individual breakdown voltages. ip

Step 1:
Find equivalent capacitance.
\[ \frac{1}{C_{\text{eq}}}=\frac{3}{C} \] \[ C_{\text{eq}}=\frac{C}{3} \] ip

Step 2:
Find breakdown voltage of the combination.
Each capacitor can tolerate maximum voltage \(V\). In series, total safe voltage is: \[ V+V+V=3V \] ip Hence, the correct answer is:
\[ \boxed{(D)\ \frac{C}{3},\ 3V} \]
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