JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Capacitance

If the charge on a capacitor is increased by 2 C, the energy stored in it increases by

44%

. The original charge on the capacitor is (in C) :

(A)10✓ Correct
(B)20
(C)30
(D)40

Answer: (A)

Solution

Energy stored in a capacitor is

U = \frac{q ^{2}}{2 C}

, so

U \propto q^{2}

.

If energy increases by 44%, then

U_{f} = 1.44 U_{i}

.

Since

U \propto q^{2}

\frac{U _{f}}{U _{i}} = \frac{q _{f}^{2}}{q _{i}^{2}} = 1.44

Therefore

q_{f} = 1.2 q_{i}

.

Given

q_{f} - q_{i} = 2

1.2 q_{i} - q_{i} = 2

0.2 q_{i} = 2

, so

q_{i} = 10

C.

Answer: (A) 10

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option

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