Question

A pure capacitor is connected to an ac power supply. In this circuit, the current

A) lags the voltage by 90 degree.
B) leads the voltage by 90 degree.
C) lags the voltage by 180 degree.
D) is in phase with the voltage.
E) None of the given answers are correct.

Answer 1

B) leads the voltage by
`90` degree.

Step-by-step explanation:

The capacitor having capacitance
`C` is connected with the AC power supply as shown in the figure.

Let the voltage of the power source be

`V= V_0 \sin (\omega t)\; \cdots (i)`,

where
`\omega` is the angular frequency,
`t` represents time and
`V_0` is the maximum voltage. Here,
`\omega` and
`V_0` is constants.

At any instant, let
`q` be the charge on the capacitor and as the potential difference across the capacitor is
`V`, so
`q=CV`.

`\Rightarrow q= CV_0 \sin (\omega t)\; \cdots (ii)`

As the current,
`i` , in the circuit is the rate of change of charge,
`q`, so

`i=\frac {dq}{dt}`

`\Rightarrow i= (d)/(dt)(CV_0 \sin (\omega t))` [from equation
`(ii)`]

`\Rightarrow i= C\omega V_0 \cos (\omega t)`

`\Rightarrow i= C\omega V_0 \cos\left(\omega t+\frac{\pi}2\right)\;\cdots(iii)`

Now, from equations
`(i)` and
`(iii)`, the phase difference between current and the voltage is
`\frac {\pi}{2}` where current is leading.

So, the current leads the voltage by
`\frac {\pi}{2}` or
`90` degree.

Hence, option
`(b)` is correct.