A sinusoidal voltage V(t) = 100 sin(500t) is | vedclass

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(B) In a pure inductive circuit,the current always lags behind the electromotive force (emf) by a phase angle of $\frac{\pi}{2}$.Given the voltage $V(

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$-10 \cos(500t)$

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(B) In a pure inductive circuit,the current always lags behind the electromotive force (emf) by a phase angle of $\frac{\pi}{2}$.Given the voltage $V(t) = V_0 \sin(\omega t)$,the current is given by $I(t) = I_0 \sin(\omega t - \frac{\pi}{2})$.Here,$V_0 = 100 \, V$ and $\omega = 500 \, rad/s$.The peak current $I_0$ is calculated as $I_0 = \frac{V_0}{\omega L} = \frac{100}{500 	imes 0.02} = \frac{100}{10} = 10 \, A$.Substituting the values into the current equation:$I(t) = 10 \sin(500t - \frac{\pi}{2})$.Using the trigonometric identity $\sin(	heta - \frac{\pi}{2}) = -\cos(	heta)$,we get:$I(t) = -10 \cos(500t)$.

$A$ sinusoidal voltage $V(t) = 100 \sin(500t)$ is applied across a pure inductance of $L = 0.02 \, H$. The current through the coil is:

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