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Physics · Electromagnetic Induction

JEE Main 2025 — 23 January, Morning Shift — Question 67

In the given circuit the sliding contact is pulled outwards such that electric current in the circuit changes at the rate of 8 A/s8 \mathrm{~A} / \mathrm{s}. At an instant when R is 12Ω12 \Omega, the value of the current in the circuit will be \qquad A.

Question figure

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

ε−LdIdt−IR=0\varepsilon-\frac{\mathrm{LdI}}{\mathrm{dt}}-\mathrm{IR}=0

12−3×(−8)−I×12=012-3 \times(-8)-\mathrm{I} \times 12=0

I=3\mathrm{I}=3

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction
Topic
L-R, L-C and L-C-R circuits with DC supply

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