Question
A small signal voltage $$V\left( t \right) = {V_0}\sin \omega t$$ is applied across an ideal capacitor $$C$$
A.
over a full cycle the capacitor $$C$$ does not consume any energy from the voltage source
B.
current $$l\left( t \right)$$ is in phase with voltage $$V\left( t \right)$$
C.
current $$l\left( t \right)$$ leads voltage $$V\left( t \right)$$ by $${180^ \circ }$$
D.
current $$l\left( t \right),$$ lags voltage $$V\left( t \right)$$ by $${90^ \circ }$$
Answer :
over a full cycle the capacitor $$C$$ does not consume any energy from the voltage source
Solution :
For an $$AC$$ circuit containing capacitor only, the phase difference between current and voltage will be $$\frac{\pi }{2}\left( {{\text{i}}{\text{.e}}{\text{.}}\,{{90}^ \circ }} \right).$$
In this case current is ahead of voltage by $$\frac{\pi }{2}.$$
Hence, power in this case is given by $$P = VI\cos \phi $$ ($$\phi = $$ phase difference between voltage and current)
$$P = VI\cos {90^ \circ } = 0$$