Question

A resistor of resistance \( R \), an inductor of inductive reactance \( 2R \), and a capacitor of capacitive reactance \( 3R \) are connected in series to an ac source. The power factor of the series LCR circuit is:

• A. \( \frac{1}{3} \)
• B. \( \frac{1}{\sqrt{3}} \)
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( \frac{1}{2} \)

Hint:

In an LCR series circuit, the power factor is crucial for understanding how the phase of the voltage compares to the phase of the current, indicating how resistive or reactive the circuit is.

Answer 1

The Correct Option is C

Step 1: Calculate the net reactance in the circuit. Since the inductive reactance (\( X_L \)) is \( 2R \) and the capacitive reactance (\( X_C \)) is \( 3R \), the net reactance \( X \) is: \[ X = X_L - X_C = 2R - 3R = -R \] The negative sign indicates that the circuit is capacitive. 
Step 2: Determine the total impedance \( Z \). \[ Z = \sqrt{R^2 + X^2} = \sqrt{R^2 + (-R)^2} = R\sqrt{2} \] 
Step 3: Calculate the power factor. The power factor is the cosine of the phase angle \( \phi \), where \( \phi \) is the angle whose tangent is the ratio of the total reactance to the resistance. 
Since \( X = -R \), we have: \[ \tan(\phi) = \frac{X}{R} = \frac{-R}{R} = -1 \] The corresponding phase angle \( \phi \) is \( -45^\circ \), and thus: \[ \cos(\phi) = \cos(-45^\circ) = \frac{1}{\sqrt{2}} \]