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  4. 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) ) :

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Q. 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 $\left( t_0 > > \frac{L}{R} \right)$ :

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Solution:

$I=\frac{v}{\omega L}\left(\omega t-\frac{\pi}{2}\right).$