QJEEM25Ph5e8e: JEE Main 2025 Physics - Class 12 JEEM25Ph5e8e Solution: In the given circuit the sliding contact is pulled outwards…

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

$\varepsilon - \frac{L \, dI}{dt} - I \cdot R = 0$

For this circuit, we have:

$\varepsilon = 12 \, \text{V}$ (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 $3 \, \text{A}$ .