JEE Main 2019 — Physics solution | Fundamenthol

JEE Main2019Physics

Q.

In the figure shown, after the switch ‘S’ is turned from position ‘A’ to position ‘B’, the energy dissipated in the circuit in terms of capacitance ‘C’ and total charge ‘Q’ is

A
$\frac{3}{8} \frac{Q ^{2}}{C}$
B
$\frac{1}{8} \frac{Q ^{2}}{C}$
C
$\frac{5}{8} \frac{Q ^{2}}{C}$
D
$\frac{3}{4} \frac{Q ^{2}}{C}$

Solution

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