JEE PYQ: Capacitance Question 15
Question 15 - 2020 (04 Sep Shift 2)
A capacitor $C$ is fully charged with voltage $V_0$. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{C}{2}$. The energy loss in the process after the charge is distributed between the two capacitors is:
(1) $\frac{1}{2}CV_0^2$
(2) $\frac{1}{3}CV_0^2$
(3) $\frac{1}{4}CV_0^2$
(4) $\frac{1}{6}CV_0^2$
Show Answer
Answer: (4)
Solution
Heat loss $H = \frac{C_1 C_2}{2(C_1 + C_2)}(V_1 - V_2)^2$. Put $C_1 = C$, $C_2 = \frac{C}{2}$, $V_1 = V_0$, $V_2 = 0$. Loss $= \frac{C \cdot \frac{C}{2}}{2(C + \frac{C}{2})}(V_0)^2 = \frac{C}{6}V_0^2$.
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