A e.m.f. $e_s=50 \sin 314t$ is applied across a pure capacitor of 637 μF. The instantaneous current I is:

$10 \cos 314t\, A$

The correct answer is Option (1) → $10 \cos 314t\, A$

The emf $(E_s)$ applied across a pure capacitor -

$E_s=E_0\sin(ωt)$

where,

$E_0$ = Amplitude of emf = 50 V

$ω$ = Angular frequency = $2\pi f=314\,rad/s$

$I(t)=C\frac{dv}{dt}$

where,

$C=637μf=637×10^{-6}F$

$V(t)=E_0\sin(ωt)$

$I=C\frac{d(E_0\sin(ωt))}{dt}=CE_0ω\cos(ωt)$

$I(t)=(637×10^{-6})×50×314×\cos(314t)$

$=10.0\cos(314t)A$