JEE PYQ: Capacitance Question 16
Question 16 - 2020 (05 Sep Shift 1)
Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is:
(1) $\frac{25}{6}CV^2$
(2) $\frac{3}{2}CV^2$
(3) zero
(4) $\frac{9}{2}CV^2$
Show Answer
Answer: (2)
Solution
$Q_1 = CV$, $Q_2 = 2C \times 2V = 4CV$. When connected in parallel with opposite polarity: By conservation of charge $4CV - CV = (C + 2C)V_{common}$. $V_{common} = \frac{3CV}{3C} = V$. Final energy $= \frac{1}{2}CV^2 + \frac{1}{2} \times 2CV^2 = \frac{3}{2}CV^2$.
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