JEE Main 2021 · 1 September · evening · Single correct · +4 / −1
For the given circuit the current i through the battery when the key in closed and the steady state has been reached is __________.
- (a) 6 A
- (b) 25 A
- (c) 10 A
- (d) 0 A
Answer and solution
Answer (c) 10 A
Solution
Physics sir · 3 steps
Once the current stops changing, every inductor is just a plain wire. Then it is ordinary resistor work.
- Inductors at steady state. An inductor opposes only a changing current, since V_L = Ldi/dt. In steady state di/dt = 0, so both inductors act as zero-resistance wires, whatever their values.
- Three equal branches. Each of the three parallel branches is now just a 3 Ω resistor. Three equal resistors in parallel give 3/3 = 1 Ω.
i = (30 V)/(2 Ω + 1 Ω) = 10 A
- So. The 2 Ω resistor is in series with the 1 Ω combination. That gives 3 Ω in total, so the battery current is 10 A, option (c).
The idea to keep. At DC steady state an inductor is a short circuit and a capacitor is an open circuit. Replace each one, then solve the resistor network.
Answer from the NTA final answer key. Solution by PrepWiser.