JEE Main 2025 — Electromagnetic Induction Question with Solution

To analyze the given circuit, we can use the equation for induced electromotive force (EMF) in an inductor. The formula is:

$\epsilon -\frac{LdI}{dt}-I\cdot R=0$

For this circuit, we have:

$\epsilon =12\text{V}$ (the applied EMF)

$L=3\text{H}$ (the inductance of the inductor)

$\frac{dI}{dt}=8\text{A/s}$ (the rate of change of current)

Assuming the direction of current is such that the change in current is negative, substitute these values into the equation:

$12-3\times (-8)-I\times 12=0$

Solving for $I$ (the current in the circuit):

$12+24-12I=0$

$36=12I$

$I=\frac{36}{12}=3\text{A}$

Thus, the current at the moment when the resistance $R$ is $12\Omega$ is $3\text{A}$.