JEE PYQ: Capacitance Question 12
Question 12 - 2020 (02 Sep Shift 1)
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 $\frac{X}{100}$ J then value of $X$ to the nearest integer is ____.
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Answer: 4
Solution
Given, $C_1 = 5,\mu$F and $V_1 = 220$ Volt. $C_2 = 2.5,\mu$F, $V_2 = 0$. Energy change: $\Delta U = U_i - U_f = \frac{1}{2}\frac{C_1 C_2}{C_1 + C_2}(V_1 - V_2)^2 = \frac{1}{2} \times \frac{5 \times 2.5}{5 + 2.5} \times (220)^2 \times 10^{-6} = \frac{5 \times 11 \times 22}{3} \times 10^{-4}$ J $= 4 \times 10^{-2}$ J. So $\frac{X}{100} = 4 \times 10^{-2}$, $X = 4$.
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