JEE Main 2026 Apr 6 Shift 1, Physics Q18: Inductance

JEE Main2026Apr 6, Shift 1Physics

Q.

An inductor of inductance $10$ mH having resistance of $100 Ω$ is connected to battery of E.M.F. $1.0$ V through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is $2$ mA and $4$ mA is _______.

A
$\frac{4}{3}$
B
$\frac{3}{4}$
C
$\frac{5}{3}$
D
$\frac{3}{5}$

Solution

By KVL around the loop: $E = i R + L \frac{d i}{d t}$ , where $L d i / d t$ is the voltage across the pure inductive element. Hence

$V_{L} = E - i R$ .

At $i_{1} = 2$ mA: $V_{L_{1}} = 1 - (2 \times 1 0^{- 3}) (100) = 1 - 0.2 = 0.8$ V.

At $i_{2} = 4$ mA: $V_{L_{2}} = 1 - (4 \times 1 0^{- 3}) (100) = 1 - 0.4 = 0.6$ V.

$\frac{V _{L_{1}}}{V _{L_{2}}} = \frac{0.8}{0.6} = \frac{4}{3}$ .

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