Question:
A series \(LR\) circuit is connected to a voltage source with \(V(t) = V_0 \sin\Omega t\). After very large time, current \(I(t)\) behaves as \(\left(t_0 \gg \frac{L}{R}\right)\):
A series \(LR\) circuit is connected to a voltage source with \(V(t) = V_0 \sin\Omega t\). After very large time, current \(I(t)\) behaves as \(\left(t_0 \gg \frac{L}{R}\right)\):
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For a long time the exponentially decaying transient term dies out and only the steady-state sinusoidal current of constant amplitude remains.
Updated On: Nov 11, 2026
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The Correct Option is B
Solution and Explanation
Equation of the circuit: \(L\frac{dI}{dt} + RI = V_0\sin\Omega t\).
Its solution is \(I(t) = \frac{V_0}{\sqrt{R^2+\Omega^2L^2}}\sin(\Omega t - \phi) + A e^{-Rt/L}\), where \(\tan\phi = \frac{\Omega L}{R}\).
The second (transient) term decays with time constant \(L/R\), so for \(t_0 \gg L/R\) it is negligible.
The current is then a sinusoid of constant amplitude \(\frac{V_0}{\sqrt{R^2+\Omega^2L^2}}\), lagging the voltage by \(\phi\). Graph (1) shows a decaying amplitude, (3) a pure decay and (4) a constant DC value, none of which is possible for an AC source at long times.
Answer: (2) a constant-amplitude sinusoidal current (graph 2)
Its solution is \(I(t) = \frac{V_0}{\sqrt{R^2+\Omega^2L^2}}\sin(\Omega t - \phi) + A e^{-Rt/L}\), where \(\tan\phi = \frac{\Omega L}{R}\).
The second (transient) term decays with time constant \(L/R\), so for \(t_0 \gg L/R\) it is negligible.
The current is then a sinusoid of constant amplitude \(\frac{V_0}{\sqrt{R^2+\Omega^2L^2}}\), lagging the voltage by \(\phi\). Graph (1) shows a decaying amplitude, (3) a pure decay and (4) a constant DC value, none of which is possible for an AC source at long times.
Answer: (2) a constant-amplitude sinusoidal current (graph 2)
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