Three capacitors each of capacitance C and of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) When $n$ capacitors of capacitance $C$ are connected in series,the equivalent capacitance $C_{	ext{eq}}$ is given by $\frac{1}{C_{	ext{eq}}} = \

Vedclass · Accepted Answer

$C/3, 3V$

Vedclass · Answer

(B) When $n$ capacitors of capacitance $C$ are connected in series,the equivalent capacitance $C_{	ext{eq}}$ is given by $\frac{1}{C_{	ext{eq}}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$.Thus,$C_{	ext{eq}} = \frac{C}{3}$.In a series combination,the potential difference across each capacitor is $V$. Since the capacitors are identical,the total breakdown voltage $V_{	ext{total}}$ of the combination is the sum of the individual breakdown voltages.$V_{	ext{total}} = V + V + V = 3V$.Therefore,the combination has a capacitance of $C/3$ and a breakdown voltage of $3V$.

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

Similar Questions

Two capacitors of same capacity are first joined in series and then in parallel. The ratio of resultant capacity in series to that in parallel combination will be

In the circuit as shown in the figure,the effective capacitance between $A$ and $B$ is ........ $\mu F$.

To obtain $3\,\mu F$ capacitance from three capacitors of $2\,\mu F$ each,they will be arranged as:

Two capacitors each having a capacitance $2 \times 10^{-6} \ F$ and a breakdown voltage $5000 \ V$,are joined in series. What will be the resultant capacitance and the breakdown voltage of the combination?

In the given circuit, the potential difference across the $4 \mu\text{F}$ capacitor is (in $text{ V}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module