A 5 μ F capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 μ F capacitor. If the energy change during the charge redistribution is X/100 J then value of X to the nearest integer is _____. Numerical answer type. Work it out, then open the answer.

Answer: 4. Joining the two capacitors in a loop shares charge: Δ U = 1/2(C₁C₂)/(C₁ + C₂)V² (C₁C₂)/(C₁ + C₂) = (5 × 2.5)/(7.5) = 5/3 μ F Δ U = 1/2 × 5/3 × 10⁻⁶ × 220² ≈ 0.0403 J = (4.03)/(100) J So X = 4.

A 5 μ F capacitor is charged fully by a 220 V supply. It is then… | JEE Main 2020 (2 Sep, Shift 1)

Q28JEE Main 2020 · 2 SEP · SHIFT 1Medium

Charge Sharing & Loss of Energy

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

Numerical answer type. Work it out, then open the answer.

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Answer: 4

Joining the two capacitors in a loop shares charge: $\Delta U = \frac{1}{2}\frac{C_1C_2}{C_1 + C_2}V^2$
$\frac{C_1C_2}{C_1 + C_2} = \frac{5 \times 2.5}{7.5} = \frac{5}{3}\ \mu$ F
$\Delta U = \frac{1}{2} \times \frac{5}{3} \times 10^{-6} \times 220^2 \approx 0.0403$ J $= \frac{4.03}{100}$ J
So $X = 4$ .

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