Question

Three capacitors are connected as shown inf figure, where $C_1=3\mu \text{F},C_2=6\mu \text{F}$, $C_3=2\mu \text{F}$ and $E=10\text{V}$. Then the


(1) Energy drawn from the source is $4\times {10}^{-4}\text{J}$
(2) Energy drawn from the source is $2\times {10}^{-4}\text{J}$
(3) Heat generated in the system during charging is $2\times {10}^{-4}\text{J}$
(4) charges on the three capacitors are equal.

Which of the above statements are correct?

• A. $1$ & $3$
• B. $2$ & $3$
• C. $1,3$ & $4$
• D. $1$ & $3$ only

Answer 1

The correct option is C $1,3$ & $4$

The equivalent capacitance of the given circuit.

$C_{eq}=\frac{3\times 6}{3+6}+2=4\mu \text{F}$

Total charge in the circuit is,

$\Delta Q=C_{eq}V=4\times 10=40\mu \text{C}$

The energy drawn from the source is,

$\Delta QV=400\mu \text{J}$

Potential energy stored in the capacitors is

$U=\frac{1}{2}{C}_{eq}{V}^2$

$U=\frac{1}{2}\times 4\times (10{)}^2=200\mu \text{J}$

Heat generated in the system during charging will be,

$\text{Work done}=H+\Delta U$

$\Delta QV=H+200$

$400=H+200$

$H=200\mu \text{J}$



$\because$ Parallel branches has equal capacitance, hence charge on them will be equal.

$\because Q=CV$

$Q=2\times 10$

$Q=20\mu \text{C}$

So the charges on the three capacitors are equal.

Hence, option $C$ is correct.