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: Write the power factor of a series L-C-R circuit.
\[ \cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^{2} + \left(\omega L - \dfrac{1}{\omega C}\right)^{2}}} \]
Step 2: Set the power factor equal to 1.
\(\cos\phi = 1\) requires \(Z = R\), which means the reactive part must vanish:
\[ \omega L - \frac{1}{\omega C} = 0 \]
Step 3: Solve the condition.
\[ \omega L = \frac{1}{\omega C} \]
This is the resonance condition, where inductive reactance equals capacitive reactance.
Step 4: Match with options. This is option (B). Option (A) is dimensionally wrong, and option (C) does not make the reactance zero.
\[\boxed{\omega L = \dfrac{1}{\omega C}}\]