JEE Main 2021 · 24 February · evening · Single correct · +4 / −1
Figure shows a circuit that contains four identical resistors with resistance R = 2.0Ω, two identical inductors with inductance L = 2.0 mH and an ideal battery with emf E = 9V. The current 'i' just after the switch 'S' is closed will be :
- (a) 3.0 A
- (b) 3.37 A
- (c) 9 A
- (d) 2.25 A
Answer and solution
Answer (d) 2.25 A
Solution
Physics sir · 4 steps
Just after closing, each inductor acts like a broken wire, so cut both L branches out.
- Inductor at start. An inductor opposes any sudden change in current. Before S is closed its current is zero, so just after closing it is still zero. Each inductor therefore acts as an open circuit.
- Drop the L branches. The two right-hand branches each contain an inductor, so both carry no current. Only the middle branch with a single R is left between the top and bottom wires.
- Series circuit. That middle R is in series with the bottom R, so the battery sees 2R = 4.0 Ω.
i = E/2R = (9 V)/(4.0 Ω) = 2.25 A
- So. Just after the switch is closed, i = 2.25 A, which is option (d).
The idea to keep. Just after switching on, an uncharged inductor is an open circuit. Long after, it is a plain wire. Redraw the circuit for the instant you are asked about, then use ordinary series and parallel rules.
Answer from the NTA final answer key. Solution by PrepWiser.