Question

A resistor of resistance \( R \), inductor of inductive reactance \( 2R \) and a capacitor of capacitive reactance \( X_C \) are connected in series to an A.C. source. If the series LCR circuit is in resonance, then the power factor of the circuit and the value \( X_C \) are respectively:

• A. \( 0.5 \) and \( 4R \)
• B. \( 1 \) and \( 2R \)
• C. \( 0.5 \) and \( 2R \)
• D. \( 1 \) and \( 4R \)

Hint:

At resonance, \( X_L = X_C \) and the power factor is always \( 1 \).

Answer 1

The Correct Option is B

Step 1: Resonance Condition At resonance, inductive reactance equals capacitive reactance: \[ X_L = X_C \] Step 2: Compute \( X_C \) \[ X_C = 2R \] Step 3: Compute Power Factor At resonance, power factor is given by: \[ \cos \phi = 1 \] Thus, the correct answer is \( 1 \) and \( 2R \).

Answer 2

Given:
- Resistance \( R \)
- Inductive reactance \( X_L = 2R \)
- Capacitive reactance \( X_C \) (unknown)

The three components \( R \), \( L \), and \( C \) are connected in series to an AC source.

Step 1: Condition for resonance in series LCR circuit:
Resonance occurs when the inductive reactance equals the capacitive reactance:
\[ X_L = X_C \] Given \( X_L = 2R \), so at resonance:
\[ X_C = 2R \]

Step 2: Power factor of the circuit at resonance:
At resonance, the inductive and capacitive reactances cancel each other, so the circuit behaves purely resistive.
Power factor \( = \cos \phi = 1 \)

Therefore,
- Power factor \( = 1 \)
- Capacitive reactance \( X_C = 2R \)

Final answer:
Power factor = 1, \( X_C = 2R \)