An e.m.f. \( e_s = 50\sin 314t \) is applied across a pure capacitor of \( 637\,\mu\mathrm{F} \). The instantaneous current I is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 10\cos 314t\,\mathrm{A} \)
A pure capacitor in an ac circuit offers a capacitive reactance \( X_C = 1/(\omega C) \), and the current through it leads the voltage by \( \pi/2 \).
\( \omega = 314\,\mathrm{rad\,s^{-1}} \)
\( C = 637\,\mu\mathrm{F} = 637\times10^{-6}\,\mathrm{F} \)
\( \displaystyle X_C = \frac{1}{\omega C} = \frac{1}{314 \times 637\times10^{-6}} = \frac{1}{0.200} = 5\,\Omega \)
\( \displaystyle I_0 = \frac{e_0}{X_C} = \frac{50\,\mathrm{V}}{5\,\Omega} = 10\,\mathrm{A} \)
\( \displaystyle I = I_0\sin\left(314t + \frac{\pi}{2}\right) = 10\cos 314t\,\mathrm{A} \)