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A sinusoidal voltage $\mathrm{V}(\mathrm{t})=100 \sin (500 \mathrm{t})$ is applied across a pure inductance of $L=0.02 \mathrm{H}$. The current through the coil is:
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The correct answer is:
$-10 \cos (500 \mathrm{t})$
$-10 \cos (500 \mathrm{t})$
In a pure inductive circuit current always lags behind the emf by $\frac{\pi}{2}$.
If $\mathrm{v}(\mathrm{t})=\mathrm{v}_0 \sin \omega \mathrm{t}$ then $I=I_0 \sin \left(\omega t-\frac{\pi}{2}\right)$ Now, given $v(t)=100 \sin (500 t)$ and $\mathrm{I}_0=\frac{\mathrm{E}_0}{\omega \mathrm{L}}=\frac{100}{500 \times 0.02}[\because \mathrm{L}=0.02 \mathrm{H}]$ $\mathrm{I}_0=10 \sin \left(500 \mathrm{t}-\frac{\pi}{2}\right)$ $\mathrm{I}_0=-10 \cos (500 \mathrm{t})$
If $\mathrm{v}(\mathrm{t})=\mathrm{v}_0 \sin \omega \mathrm{t}$ then $I=I_0 \sin \left(\omega t-\frac{\pi}{2}\right)$ Now, given $v(t)=100 \sin (500 t)$ and $\mathrm{I}_0=\frac{\mathrm{E}_0}{\omega \mathrm{L}}=\frac{100}{500 \times 0.02}[\because \mathrm{L}=0.02 \mathrm{H}]$ $\mathrm{I}_0=10 \sin \left(500 \mathrm{t}-\frac{\pi}{2}\right)$ $\mathrm{I}_0=-10 \cos (500 \mathrm{t})$
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