Question

In an AC circuit a resistor \(100\Omega\), inductor \(0.1\,\text{mH}\) and a capacitor are connected in series across an AC source of \(220\,\text{V},\,50\,\text{Hz}\). If the power factor of the circuit is \( \frac{1}{2} \) and \( |X_L - X_C| = \alpha\sqrt{3}\,\Omega \), find \(\alpha\).

• A. \(100\sqrt{3}\)
• B. \(100\)
• C. \( \frac{100}{\sqrt{3}} \)
• D. \(1000\)

Answer 1

The Correct Option is B

Concept:
For a series AC circuit, the power factor is \[ \cos\phi = \frac{R}{Z} \] where \(Z\) is the impedance. Also, \[ \tan\phi = \frac{|X_L - X_C|}{R} \] These relations allow determination of the net reactance. Step 1: Use the given power factor. \[ \cos\phi = \frac{1}{2} \] \[ \phi = 60^\circ \] Step 2: Use the tangent relation. \[ \tan\phi = \frac{|X_L - X_C|}{R} \] \[ \tan 60^\circ = \frac{|X_L - X_C|}{100} \] \[ \sqrt{3} = \frac{|X_L - X_C|}{100} \] Step 3: Calculate net reactance. \[ |X_L - X_C| = 100\sqrt{3} \] Given \[ |X_L - X_C| = \alpha\sqrt{3} \] Thus, \[ \alpha = 100 \]