Be reduced by half
Be doubled
Be four times at high
Be reduced to one fourth
Answer is Right!
Answer is Wrong!
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.