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
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
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Capacitive reactance decreases when frequency increases:
\[
X_C=\frac{1}{2\pi f C}
\]
Doubling frequency halves the reactance.
Updated On: Jun 22, 2026
- \(1.5\;k\Omega\)
- \(3\;k\Omega\)
- \(6\;k\Omega\)
- \(5.2\;k\Omega\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the formula for capacitive reactance.
Capacitive reactance is given by \[ X_C=\frac{1}{2\pi f C} \] where \[ f=\text{frequency} \] and \[ C=\text{capacitance} \]
Step 2: Understand the relation with frequency.
From the formula, \[ X_C\propto \frac{1}{f} \] Thus, capacitive reactance is inversely proportional to frequency.
Step 3: Apply the condition of double frequency.
If frequency becomes double, \[ f'=2f \] then new reactance becomes \[ X_C'=\frac{X_C}{2} \] Given, \[ X_C=3\;k\Omega \] Therefore, \[ X_C'=\frac{3}{2} \] \[ X_C'=1.5\;k\Omega \]
Step 4: Final conclusion.
Hence, the new capacitive reactance is \[ \boxed{1.5\;k\Omega} \]
Capacitive reactance is given by \[ X_C=\frac{1}{2\pi f C} \] where \[ f=\text{frequency} \] and \[ C=\text{capacitance} \]
Step 2: Understand the relation with frequency.
From the formula, \[ X_C\propto \frac{1}{f} \] Thus, capacitive reactance is inversely proportional to frequency.
Step 3: Apply the condition of double frequency.
If frequency becomes double, \[ f'=2f \] then new reactance becomes \[ X_C'=\frac{X_C}{2} \] Given, \[ X_C=3\;k\Omega \] Therefore, \[ X_C'=\frac{3}{2} \] \[ X_C'=1.5\;k\Omega \]
Step 4: Final conclusion.
Hence, the new capacitive reactance is \[ \boxed{1.5\;k\Omega} \]
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