Figure shows a circuit that contains four identical resistors with resistance $\mathrm{R}=2.0 \Omega$, two identical inductors with inductance $\mathrm{L}=2.0 \mathrm{mH}$ and an ideal battery with emf $\mathrm{E}=9 \mathrm{~V}$. The current ' i ' just after switch ' $S$ ' is closed will be:

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Electromagnetic Induction

Figure shows a circuit that contains four identical resistors with resistance

R = 2.0Ω

, two identical inductors with inductance

L = 2.0 mH

and an ideal battery with emf

E = 9 V

. The current ' i ' just after switch '

S

' is closed will be:

(A)2.25 A✓ Correct
(B)3.0 A
(C)9 A
(D)3.37 A

Answer: (A)

Solution

The current through an inductor cannot change instantaneously, so immediately after the switch is closed both inductive branches still carry zero current and act as open circuits. Only the branch containing a bare resistor conducts, and it is in series with the resistor in the battery loop, giving

R_{eff} = R + R = 2 (2.0 Ω) = 4.0 Ω

The ammeter therefore reads

i = \frac{E}{R _{eff}} = \frac{9}{4} = 2.25

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option

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