JEE Main 2022 · 28 July · evening · Numerical answer · +4 / −0
For the given circuit the current through battery of 6 V just after closing the switch 'S' will be _________ A.
Answer and solution
Answer 1
Solution
Physics sir · 3 steps
At the instant the switch closes, the inductor blocks any sudden current, so treat that branch as cut.
- Inductor at start. Before the switch closes, no current flows through the 2 H coil. An inductor resists any sudden change in current, so just after closing its current is still zero. It behaves like an open switch.
- What is left. The right branch (coil, 2 Ω, 2 Ω) carries no current. The battery now drives current only through the 2 Ω on the left and the 4 Ω in the middle, which are in series.
I = (6 V)/(2 Ω + 4 Ω) = 1 A
- So. The current through the 6 V battery just after closing S is 1 A.
The common slip: 1.5. Treating the inductor as a plain wire puts the 4 Ω right branch in parallel with the 4 Ω middle branch, giving 6/4 = 1.5 A. That is the steady current long after closing, not the current just after.
The idea to keep. Just after switching on, an uncharged inductor is an open circuit. Long after, it is a plain wire. Decide which instant the question asks about before you redraw the circuit.
Answer from the NTA final answer key. Solution by PrepWiser.