JEE Physics 2023 Question with Solution

Given: Capacitance of first capacitor $C = 900 \, \mu \text{F} = 900 \times 10^{-6} \, \text{F}$, initial potential difference $V = 100 \, \text{V}$. Another identical capacitor is initially uncharged.

Find: The value of $x$ if energy lost is $x \times 10^{-2} \, \text{J}$.

When a charged capacitor is connected to an identical uncharged capacitor, charge redistributes equally. Hence the final common potential becomes

$$V_f = \frac{V}{2} = 50 \, \text{V}$$

Initial energy stored in the charged capacitor is

$$U_i = \frac{1}{2} C V^2$$

Substituting the values,

$$U_i = \frac{1}{2} \times 900 \times 10^{-6} \times (100)^2 = 4.5 \, \text{J}$$

Finally, both identical capacitors together have equivalent capacitance $2C$ at common voltage $V_f = 50 \, \text{V}$. So final energy is

$$U_f = \frac{1}{2} (2C) V_f^2$$ $$U_f = \frac{1}{2} \times 2C \times \left(\frac{V}{2}\right)^2 = \frac{CV^2}{4}$$ $$U_f = \frac{900 \times 10^{-6} \times (100)^2}{4} = 2.25 \, \text{J}$$

Therefore, loss of energy is

$$\Delta U = U_i - U_f = 4.5 - 2.25 = 2.25 \, \text{J}$$

Given that

$$\Delta U = x \times 10^{-2} \, \text{J}$$

So,

$$x \times 10^{-2} = 2.25$$ $$x = 225$$

Therefore, the required value is $225$.

The solution is unrelated to this capacitor question, so the answer has been derived from the question data and the given correct answer field.