Question 441: Electrostatics & Capacitance

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

Accepted Answer

Step 1: Equivalent capacitance (for series)
For capacitors in series:
$ \frac{1}{C_{	ext{eq}}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} $
So,
$ C_{	ext{eq}} = \frac{C}{3}. $
Step 2: Breakdown voltage
Each capacitor can withstand a maximum voltage of $V$ before breakdown.
When three are in
series
, the total applied voltage is divided (equally, since they have the same capacitance). Thus, each capacitor gets $	frac{V_	ext{total}}{3}$.
For safe operation, none should exceed $V$.
So the maximum total breakdown voltage is:
$ V_{	ext{eq}} = 3V $
Final Answer:
$ C_{	ext{eq}} = \frac{C}{3}, \quad V_{	ext{eq}} = 3V $
Ã¢ÂÂ Correct option:
D

Answer

$3 \mathrm{C}, 3 \mathrm{~V}$

Answer

$\frac{\mathrm{C}}{3}, \frac{\mathrm{~V}}{3}$

Answer

$3 \mathrm{C}, \frac{\mathrm{V}}{3}$

Answer

$\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$

Electrostatics & Capacitance

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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