Question:
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
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: Apr 26, 2026
- \(0.4\)
- \(0.8\)
- \(0.6\)
- \(0.9\)
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The Correct Option is B
Solution and Explanation
Concept:
Power factor: \[ \cos\phi = \frac{R}{Z} \] where \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Step 1: Calculate net reactance. \[ X = X_L - X_C = 70 - 130 = -60\Omega \]
Step 2: Impedance. \[ Z = \sqrt{80^2 + (-60)^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\Omega \]
Step 3: Power factor. \[ \cos\phi = \frac{80}{100} = 0.8 \]
Step 4: Conclusion. Power factor = \(0.8\)
Power factor: \[ \cos\phi = \frac{R}{Z} \] where \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Step 1: Calculate net reactance. \[ X = X_L - X_C = 70 - 130 = -60\Omega \]
Step 2: Impedance. \[ Z = \sqrt{80^2 + (-60)^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\Omega \]
Step 3: Power factor. \[ \cos\phi = \frac{80}{100} = 0.8 \]
Step 4: Conclusion. Power factor = \(0.8\)
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