Three capacitors $C_1$, $C_2$ and $C_3$ are connected to a voltage source $V$ as shown in the figure. The voltage across $C_3$ will be:

  • A
    $\frac{C_3V}{(C_1 + C_2 + C_3)}$
  • B
    $\frac{(C_1 + C_2)V}{C_3}$
  • C
    $\frac{(C_2 + C_3)V}{C_1 + C_2}$
  • D
    $\frac{(C_1 + C_2)V}{(C_1 + C_2 + C_3)}$

Explore More

Similar Questions

$A$ number of capacitors,each of capacitance $1\,\mu F$ and each of which gets punctured if a potential difference exceeding $500\,V$ is applied,are provided. An arrangement suitable for giving a capacitance of $2\,\mu F$ across which $3000\,V$ may be applied requires at least:

Two capacitors having capacitances $8 \mu F$ and $16 \mu F$ have breaking voltages $20 \ V$ and $80 \ V$ respectively. They are connected in series. What is the maximum charge they can store in this combination in $\mu C$?

The equivalent capacitance between $A$ and $B$ in the figure is $1\,\mu F$. Then the value of capacitance $C$ is.....$\mu F$

Five capacitors,each of capacity $C$,are connected as shown in the figure. The resultant capacity between $A$ and $B$ is $14 \mu F$. The capacity of each capacitor is (in $\mu F$)

Three capacitors each of $6\,\mu F$ are available. The minimum and maximum capacitances which may be obtained are

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo