Question

In a purely capacitive circuit, if the supply frequency is reduced to \( \frac{1}{4} \), the current will :

• A. be reduced by half
• B. be doubled
• C. be reduced to one-fourth
• D. be four times higher

Hint:

For a purely capacitive circuit: \[ X_C=\frac{1}{2\pi f C} \] and \[ I\propto f \] So if frequency decreases, current also decreases proportionally.

Answer 1

The Correct Option is C

Concept: In a purely capacitive AC circuit, capacitive reactance is: \[ X_C=\frac{1}{2\pi f C} \] Current in the circuit is: \[ I=\frac{V}{X_C} \] Substituting the value of \(X_C\): \[ I=V(2\pi f C) \] Thus: \[ I\propto f \] Therefore current is directly proportional to frequency in a purely capacitive circuit.

Step 1: Writing the proportionality relation. Since: \[ I\propto f \] we can write: \[ \frac{I_2}{I_1}=\frac{f_2}{f_1} \]

Step 2: Using the changed frequency condition. The frequency is reduced to: \[ \frac{1}{4} \] of its original value. Thus: \[ f_2=\frac{f_1}{4} \] Substituting: \[ \frac{I_2}{I_1}=\frac{f_1/4}{f_1} \] \[ \frac{I_2}{I_1}=\frac{1}{4} \]

Step 3: Finding the new current. Therefore: \[ I_2=\frac{I_1}{4} \] Hence the current becomes one-fourth of its original value.

Step 4: Selecting the correct option. Thus the correct answer is: \[ \boxed{(3)\text{ be reduced to one-fourth}} \]