Q 12-06-041JEE MainJEE Main 2025 (23 Jan, Shift 1)Medium
In the given circuit, the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of $8\ \text{A/s}$. At an instant when $R$ is $12\ \Omega$, the value of the current in the circuit will be ______ A.
Numerical value type. Enter your answer.
Answer: 3
Pulling the slider outwards increases $R$, so the current is decreasing: $\dfrac{dI}{dt} = -8\ \text{A/s}$.
Kirchhoff's loop law with the inductor's back emf:
$$E - L\frac{dI}{dt} = IR$$
$$12 - 3(-8) = 12\,I \Rightarrow 36 = 12\,I \Rightarrow I = 3\ \text{A}$$
(The decreasing current induces an emf in the inductor that tries to keep the current flowing, adding to the cell's emf.)
Solution by Sreeraj P, M.Sc Physics