In the figure shown, a circuit contains two identical resistors with resistance $\mathrm{R}=5 \Omega$ and an inductance with $\mathrm{L}=2 \mathrm{mH}$. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed?

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Electromagnetic Induction

In the figure shown, a circuit contains two identical resistors with resistance

R = 5Ω

and an inductance with

L = 2 mH

. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed?

(A)5.5 A
(B)7.5 A
(C)3 A
(D)6 A✓ Correct

Answer: (D)

Solution

Long after the switch closes the current is steady, so the inductor no longer opposes anything and behaves like an ideal wire of zero resistance. That short circuit places the two

5 Ω

resistors in parallel, giving

R_{eq} = \frac{5 \times 5}{5 + 5} = 2.5 Ω

. The battery then drives a current

i = \frac{15}{2.5} = 6 A

through itself.

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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