Question

In L–C–R circuit power factor will be 1, if:

• A. \(\omega L = \omega C\)
• B. \(\omega L = \dfrac{1}{\omega C}\)
• C. \(\left(\omega L - \dfrac{1}{\omega C}\right) = R\)
• D. None of these

Hint:

Power factor is unity at resonance, when inductive reactance equals capacitive reactance.

Answer 1

The Correct Option is B

Step 1: Use the phasor picture of impedance.
The net reactance is \(X = X_L - X_C = \omega L - \dfrac{1}{\omega C}\), and the phase angle satisfies \(\tan\phi = X/R\).
Step 2: Power factor is \(\cos\phi\); it equals 1 only when \(\phi = 0\).
\(\phi = 0\) means \(\tan\phi = 0\), hence \(X = 0\).
Step 3: Apply \(X = 0\).
\(\omega L - \dfrac{1}{\omega C} = 0 \Rightarrow \omega L = \dfrac{1}{\omega C}\).
Step 4: Physically the circuit is purely resistive at resonance, so all supplied power is real and the power factor is unity.
\[\boxed{\omega L = \dfrac{1}{\omega C}}\]