Question

A $5\mu \text{F}$ capacitor is charged fully by a $220\text{V}$ supply. It is then disconnected from the supply and is connected across another uncharged $2.5\mu \text{F}$ capacitor. If the energy change during the charge redistribution is $\frac{X}{100}\text{J}$ then the value of $X$ to the nearest integer is

• A. 4
• B. 4.00
• C. 4.0

Answer 1

Answer: (A), (B), (C)

Given, $C_1=5\mu \text{F}\text{and}{V}_1=220\text{V}$

When capacitor $C_1$ fully charged, it is disconnected from the supply and connected to uncharged capacitor $C_2$,

$C_2=2.5\mu \text{F},V_2=0$

Energy change during the charge redistribution,

$\Delta U=U_i-U_f=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}(V_1-V_2{)}^2$

$\Rightarrow \Delta U=\frac{1}{2}\times \frac{5\times 2.5}{(5+2.5)}(220-0{)}^2\mu \text{J}$

$\Rightarrow \Delta U=\frac{5}{2\times 3}\times 22\times 22\times 100\times {10}^{-6}\text{J}$

$\Rightarrow \Delta U=\frac{1210}{3}\times {10}^{-4}\text{J}$

$\Rightarrow \Delta U\approx 4\times {10}^{-2}\text{J}$

According to question, $\frac{X}{100}=4\times {10}^{-2}$

$\therefore X=4$

Hence, $4$ is the correct answer.