JEE Main 2025 Jan 23 Shift 1, Physics Q25: Inductance

JEE Main2025Jan 23, Shift 1Physics

In the given circuit the sliding contact is pulled outwards such that electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 $Ω$ , the value of the current in the circuit will be ____ A.

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Solution

Apply Kirchhoff's voltage law around the loop. The inductor's back-emf is $ε_{L} = - L \frac{d I}{d t}$ , but as the resistance is being pulled outwards (increasing), the current is decreasing, so $\frac{d I}{d t} = - 8$ A/s.

$ε - L \frac{d I}{d t} - I R = 0$

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

$12 + 24 = 12 I \Rightarrow I = 3$ A.

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JEE Main 2025 Jan 23 Shift 1, Physics Q25: Inductance

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