Three identical resistors with resistance R = 12Ω and two identical inductors with self inductance L = 5 mH are connected to an ideal battery with emf of 12 V as shown in figure. The current through the battery long after the switch has been closed will be _____________ A.

Solution

Physics sir · 3 steps

Long after closing the switch, an ideal inductor is just a plain wire, so only the resistors matter.

1. Steady state. Long after the switch closes, the current stops changing, so Ldi/dt = 0. Each ideal inductor then has no voltage across it and acts like a plain wire.
2. Read the circuit. All three branches connect across the same two vertical wires, so they are in parallel with the battery. With the inductors replaced by wires, each branch is just one resistor of 12 Ω.

   R_(eq) = R/3 = 12/3 = 4 Ω

3. So. The battery current is I = (12 V)/(4 Ω) = 3 A, that is 3 A, with 1 A in each branch.

The common slip: 1. Treating the inductors as open circuits leaves only the middle resistor in the circuit, giving 1 A. That is the current at the instant the switch closes, not long after.

The idea to keep. Right after closing the switch, an inductor blocks a sudden change in current, so it acts as an open circuit. Long after, it carries steady current and acts as a short wire. Ask which moment the question means.

Answer from the NTA final answer key. Solution by PrepWiser.