The switch is thrown to position 1. What will be the current in the circuit in the steady state condition ?
2 A
In the steady state an inductor is a short circuit, so the only element left limiting the current is the 5 Ω series resistor.
\(I=\dfrac{V}{R}=\dfrac{10}{5}=2\ \text{A}\)
— option 2.
Why the inductor disappears from the calculation. The voltage across an inductor is
\(v_{L}=L\dfrac{di}{dt}\)
and "steady state" means precisely that the current has stopped changing, so \(di/dt=0\) and \(v_{L}=0\). A component with no voltage across it while carrying current is behaving as a short circuit. The 1 H value therefore does not appear in the answer at all — it governs only how long the approach takes, not where it ends.
The full transient, for context. Immediately after the switch closes the inductor opposes any change of current, so \(i(0)=0\), and the current then rises exponentially:
\(i(t)=\dfrac{V}{R}\left(1-e^{-t/\tau}\right),\qquad \tau=\dfrac{L}{R}=\dfrac{1}{5}=0.2\ \text{s}\)
After one time constant the current has reached 63 % of its final value, and after five it is within 1 % — so the steady state here is effectively reached in about a second.
| Element | At t = 0+ | At steady state |
|---|---|---|
| Inductor | Open circuit | Short circuit |
| Capacitor | Short circuit | Open circuit |
Why the second 5 Ω resistor is irrelevant. It lies on the branch reached through contact 2, which the pole is not touching. Nothing connects it into the loop, so no current flows through it. Its role is to give the inductor a discharge path when the switch is later thrown to position 2 — without it, breaking the current in an inductor would generate a large \(L\,di/dt\) spike across the opening contacts. Including it in the sum, as the distractor 1 A would require, is the error the question is set to catch.
Hence, the steady-state current is 2 A.
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