Question

If AC voltage is applied to a pure capacitor, then voltage across the capacitor...
(A) Leads the current by phase angle $ \dfrac{\pi }{2} $ rad
(B) Leads the current by phase angle $ \pi $ rad
(C) Lags behind the current by phase angle $ \dfrac{\pi }{2} $ rad
(D) Lags behind the current by phase $ \pi $ rad

Answer 1

Hint : When the ac voltage is first applied, at the first cycle the current flows into the capacitor and hence charges it. When the capacitor is charged at the peak voltage of the ac electricity the current stops flowing in the capacitor.

Complete step by step answer:
We are told to determine what happens to the voltage of ac electricity when it is applied to a pure capacitor. The phrase “applied to a pure capacitor” signifies that there are no resistive elements or inductive elements in the circuit.
Now, in such cases when ac voltage is applied to a capacitor, during the first cycle as the ac voltage increases, current flows into the capacitor. When the voltage of the ac electricity gets to the peak the currents stop flowing into the capacitor. As the voltage decreases, the current starts to flow out of the capacitor (current flowing in reverse direction) and this occurs until the voltage gets to the negative peak value. At this time, the capacitor would have been charged with the exact opposite polarity as it first had in the first cycle. We need to observe that current is max when voltage is min and voltage is max when current is mean, and since the voltage was zero at the beginning when current was high we say that the voltage lags behind the current by a quarter cycle i.e. 90 degrees.
Hence, the correct option is C.

Note:
For clarity, current reduces as voltage increases because capacitors do not allow charges to flow across it. Hence, as the charges build up at the plates (increase potential difference), they repel further charges approaching the plates. Hence, when the voltage gets to maximum, the current reaches zero.