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

• A. $3C, 3V$
• B. $\frac{C}{3}, \frac{V}{3}$
• C. $3C, \frac{V}{3}$
• D. $\frac{C}{3}, 3V$

Hint:

For \(n\) identical capacitors in series: \[ C_{\text{eq}}=\frac{C}{n} \] and total breakdown voltage becomes \(nV\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When capacitors are connected in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances.
The total breakdown voltage of the series combination is the sum of the individual breakdown voltages because the applied voltage is distributed across the capacitors.
Step 2: Key Formula or Approach:
For $n$ identical capacitors in series:
1. Equivalent Capacitance: $C_{eq} = \frac{C}{n}$
2. Total Breakdown Voltage: $V_{net} = n \times V$
Step 3: Detailed Explanation:
Given $n = 3$ identical capacitors.
1. Capacitance calculation:
\[ \frac{1}{C_{series}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \implies C_{series} = \frac{C}{3} \]
2. Breakdown voltage calculation:
Since each capacitor can withstand a maximum of $V$ volts, and they are in series, the total voltage that can be applied to the system without damaging any capacitor is the sum:
\[ V_{net} = V + V + V = 3V \]
Step 4: Final Answer:
The equivalent capacitance is $\frac{C}{3}$ and the total breakdown voltage is $3V$.