Question:medium

A series LCR circuit is connected to an a.c. source of \(230\text{ V}, 50\text{ Hz}\). The circuit contains resistance of \(80\Omega\), inductive reactance \(70\Omega\) and capacitive reactance \(130\Omega\). The power factor of the circuit is \(x\). The value of \(x\) is

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Always use net reactance \(X_L - X_C\) in impedance.
Updated On: May 14, 2026
  • \(0.4\)
  • \(0.8\)
  • \(0.6\)
  • \(0.9\)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The power factor of an AC circuit is defined as the cosine of the phase angle between voltage and current.
In a series LCR circuit, it can be calculated as the ratio of true resistance to the total impedance.
Step 2: Key Formulas or Approach:
Net reactance: \( X = X_L \sim X_C \).
Impedance of the LCR series circuit: \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).
Power factor: \( \cos\phi = \frac{R}{Z} \).
Step 3: Detailed Explanation:
The given parameters of the circuit are:
Resistance \( R = 80\Omega \)
Inductive reactance \( X_L = 70\Omega \)
Capacitive reactance \( X_C = 130\Omega \)
First, calculate the net reactance.
Since \( X_C>X_L \), the circuit is capacitive.
Net reactance \( X = |X_L - X_C| = |70\Omega - 130\Omega| = |-60\Omega| = 60\Omega \).
Next, calculate the total impedance \( Z \) of the circuit:
\[ Z = \sqrt{R^2 + X^2} \] \[ Z = \sqrt{80^2 + 60^2} \] \[ Z = \sqrt{6400 + 3600} \] \[ Z = \sqrt{10000} \] \[ Z = 100\Omega \] Now, calculate the power factor (\( \cos\phi \)), which is given as \( x \):
\[ x = \cos\phi = \frac{R}{Z} \] Substitute the values of \( R \) and \( Z \):
\[ x = \frac{80}{100} \] \[ x = 0.8 \] The values for voltage (\( 230\text{ V} \)) and frequency (\( 50\text{ Hz} \)) are not needed as the reactances are already provided directly.
Step 4: Final Answer:
The value of the power factor \( x \) is \( 0.8 \).
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