In a pure capacitive circuit if the supply frequency is reduced to $$\frac{1}{2},\,$$ the current will

The correct answer is: A. Be reduced by half.

The current in a pure capacitive circuit is given by the following equation:

$$I = \frac{V}{X_C} = \frac{V}{\frac{1}{2\pi f C}}$$

where $V$ is the voltage, $f$ is the frequency, and $C$ is the capacitance.

If the frequency is reduced to $\frac{1}{2}$, then the current will be reduced by half.

Option B is incorrect because the current will not be doubled.

Option C is incorrect because the current will not be four times at high frequency.

Option D is incorrect because the current will not be reduced to one fourth.