Question

Capacitive reactance of a capacitor in an AC circuit is \(3\;k\Omega\). If this capacitor is connected to a new AC source of double frequency, the capacitive reactance will become

• A. \(1.5\;k\Omega\)
• B. \(3\;k\Omega\)
• C. \(6\;k\Omega\)
• D. \(5.2\;k\Omega\)

Hint:

Capacitive reactance decreases when frequency increases: \[ X_C=\frac{1}{2\pi f C} \] Doubling frequency halves the reactance.

Answer 1

The Correct Option is A

Step 1: Write the capacitive reactance formula.
The opposition a capacitor offers in an AC circuit is \[ X_C = \frac{1}{2\pi f C} \] where $f$ is the frequency and $C$ the capacitance.
Step 2: Read off the dependence on frequency.
Since $C$ is fixed, the formula shows \[ X_C \propto \frac{1}{f} \] so reactance falls when frequency rises.
Step 3: Set up the ratio for the two sources.
Comparing the new and old reactances, \[ \frac{X_C'}{X_C} = \frac{f}{f'} \]
Step 4: Apply the doubling of frequency.
With $f' = 2f$, \[ \frac{X_C'}{X_C} = \frac{f}{2f} = \frac{1}{2} \]
Step 5: Compute the new reactance.
Given $X_C = 3\ \text{k}\Omega$, \[ X_C' = \frac{3}{2} = 1.5\ \text{k}\Omega \]
Step 6: State the answer.
Hence the new capacitive reactance is \[ \boxed{1.5\ \text{k}\Omega} \]