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For an LCR series circuit with an ac sou...

For an LCR series circuit with an ac source of angular frequency `omega`.

A

circuit will be capacitive if `omega gt 1/sqrt(LC)`

B

circuit will be inductive if `omega =1/sqrt(LC)`

C

power factor of circuit will by unity if capacitive reactance equals inductive reactance

D

current will be leading voltage if `omega gt 1/sqrt(LC)`

Text Solution

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The correct Answer is:
To solve the problem regarding the LCR series circuit with an AC source of angular frequency \( \omega \), we will analyze the conditions for the circuit to be capacitive, resistive, or inductive, and determine the correct statements based on these conditions. ### Step-by-Step Solution: 1. **Understanding the Circuit Behavior**: - In an LCR series circuit, the behavior depends on the relationship between the angular frequency \( \omega \), inductance \( L \), and capacitance \( C \). - The reactance of the inductor \( X_L \) is given by \( X_L = \omega L \). - The reactance of the capacitor \( X_C \) is given by \( X_C = \frac{1}{\omega C} \). 2. **Condition for Inductive Circuit**: - The circuit will be inductive if the inductive reactance is greater than the capacitive reactance: \[ \omega L > \frac{1}{\omega C} \] - Rearranging gives: \[ \omega^2 > \frac{1}{LC} \] - Thus, the circuit is inductive if: \[ \omega > \frac{1}{\sqrt{LC}} \] 3. **Condition for Resonance (Resistive Circuit)**: - The circuit is at resonance (purely resistive) when: \[ \omega = \frac{1}{\sqrt{LC}} \] - At this point, the inductive and capacitive reactances are equal, and the circuit behaves like a resistor. 4. **Condition for Capacitive Circuit**: - The circuit will be capacitive if: \[ \omega < \frac{1}{\sqrt{LC}} \] 5. **Power Factor**: - The power factor \( \cos \phi \) is defined as: \[ \cos \phi = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}} \] - At resonance, where \( X_L = X_C \), the power factor becomes: \[ \cos \phi = \frac{R}{\sqrt{R^2}} = 1 \] - This indicates that the power factor is unity at resonance. 6. **Conclusion**: - Based on the conditions derived: - Option A: Incorrect (inductive condition). - Option B: Incorrect (resistive condition). - Option C: Correct (power factor is unity at resonance). - Option D: Incorrect (current lags behind voltage in inductive circuits). ### Final Answer: The correct option is **C**: The power factor of the circuit will be unity if the capacitive reactance equals the inductive reactance.

To solve the problem regarding the LCR series circuit with an AC source of angular frequency \( \omega \), we will analyze the conditions for the circuit to be capacitive, resistive, or inductive, and determine the correct statements based on these conditions. ### Step-by-Step Solution: 1. **Understanding the Circuit Behavior**: - In an LCR series circuit, the behavior depends on the relationship between the angular frequency \( \omega \), inductance \( L \), and capacitance \( C \). - The reactance of the inductor \( X_L \) is given by \( X_L = \omega L \). - The reactance of the capacitor \( X_C \) is given by \( X_C = \frac{1}{\omega C} \). ...
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