If V = Vmsinωt is the voltage across the capacitor, then the current is:
A. $${I_m}\sin \left( {\omega t + \frac{\pi }{2}} \right)$$
B. $$\frac{{{I_m}}}{{\sqrt 2 }}$$
C. $${I_m}\sin \omega t$$
D. $${I_m}\sin \left( {\omega t - \frac{\pi }{2}} \right)$$
Answer: Option A

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