Question

A capacitor $0.2\mu \text{F}$ is charged to $600\text{V}.$ After removing the battery, it is connected to $1\mu \text{F}$ capacitor in parallel. The voltage across $1\mu \text{F}$ will become

• A. $300\text{V}$
• B. $600\text{V}$
• C. $100\text{V}$
• D. $120\text{V}$

Answer 1

The correct option is C $100\text{V}$

Given:
$C_1=0.2\mu \text{F};C_2=1\mu \text{F};V=600\text{V}$

When $C_1$ is connected to $600\text{V}$ battery, charge on $C_1$ will be $Q=C_{1}V$.

Now, the potential difference across both the capacitors will become equal after they are connected. Let it be $V^{\prime }.$

Also, the charge present on the $0.2\mu \text{F}$ capacitor initially will get distributed across both the capacitors.

Applying conservation of charge

$Q=Q_1+Q_2$

$\Rightarrow C_{1}V=C_1{V}^{\prime }+C_2{V}^{\prime }$

$\Rightarrow V^{\prime }=\frac{C_{1}V}{(C_1+C_2)}$

$\Rightarrow V^{\prime }=\frac{0.2\times 600}{0.2+1}=\frac{120}{1.2}=100\text{V}$

Hence, option $\left(c\right)$ is correct.