A series LR circuit is connected to a voltage source with V(t) = V0 sin Ω t. After very large time, current I(t) behaves as (t0 >> L/R): (a) (b) (c) (d)

Answer: (b). For t L/R the transient term e-Rt/L has died out. Only the steady-state current remains: I = (V₀)/(√(R² + Ω² L²))sin (Ω t - φ), a pure sinusoid of constant amplitude. So the graph is a sine wave of constant amplitude: answer (b).

A series LR circuit is connected to a voltage source with V(t) = V0… | JEE Main 2016 (9 Apr, Shift 1)

Q50JEE Main 2016 · 9 APR · SHIFT 1Difficult

RL and RC Series Circuits

A series LR circuit is connected to a voltage source with

V(t) = V0 sin $\Omega$ t. After very large time, current I(t) behaves as

(t0 >> $\frac{L}{R}$ ):

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Answer: (b)

For $t \gg L/R$ the transient term $e^{-Rt/L}$ has died out.
Only the steady-state current remains: $I = \frac{V_0}{\sqrt{R^2 + \Omega^2 L^2}}\sin(\Omega t - \phi)$ , a pure sinusoid of constant amplitude.
So the graph is a sine wave of constant amplitude: answer (b).

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RL and RC Series Circuits question from JEE Main 2016 (9 Apr, Shift 1)

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