Q 12-06-032JEE MainJEE Main 2026 (6 Apr, Shift 1)Medium
An inductor of inductance $10$ mH having resistance of $100\ \Omega$ 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 ______.
Answer: (A) $4/3$
The coil's own resistance takes $IR$; the rest of the emf appears across the inductance: $V_L = E - IR$.
$I = 2$ mA: $V_L = 1 - 0.2 = 0.8$ V. $\quad I = 4$ mA: $V_L = 1 - 0.4 = 0.6$ V.
Ratio $= \dfrac{0.8}{0.6} = \dfrac{4}{3}$.
Solution by Sreeraj P, M.Sc Physics