Alternating Current

JEE Main level, single correct, medium

49 of 190

Question 49 of 190

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

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Answer: (B) −10cos⁡(500t)-10 \cos (500 \mathrm{t})

In a pure inductive circuit current always lags behind the emf by π2\frac{\pi}{2}.
If v(t)=v0sin⁡ωt\mathrm{v}(\mathrm{t})=\mathrm{v}_0 \sin \omega \mathrm{t} then I=I0sin⁡(ωt−π2)I=I_0 \sin \left(\omega t-\frac{\pi}{2}\right) Now, given v(t)=100sin⁡(500t)v(t)=100 \sin (500 t) and I0=E0ωL=100500×0.02[∵L=0.02H]\mathrm{I}_0=\frac{\mathrm{E}_0}{\omega \mathrm{L}}=\frac{100}{500 \times 0.02}[\because \mathrm{L}=0.02 \mathrm{H}] I0=10sin⁡(500t−π2)\mathrm{I}_0=10 \sin \left(500 \mathrm{t}-\frac{\pi}{2}\right) I0=−10cos⁡(500t)\mathrm{I}_0=-10 \cos (500 \mathrm{t})

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