Question

In an LCR series circuit (of inductance L, capacitance C and resistance R), the impedance is minimum when the angular frequency of the source is given by

• A. \( \sqrt{LC} \)
• B. \( \frac{1}{\sqrt{LC}} \)
• C. \( \sqrt{\frac{L}{C}} \)
• D. \( \sqrt{\frac{C}{L}} \)
• E. \( \sqrt{LCR} \)

Hint:

At resonance, the circuit is purely resistive, meaning current and voltage are in phase.

Answer 1

The Correct Option is B

Concept: In a series LCR circuit, the total opposition to the flow of alternating current is known as Impedance (\(Z\)). The impedance is a combination of the ohmic resistance (\(R\)) and the net reactance (\(X_L - X_C\)). The general expression for impedance is: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Where:
• \( X_L = \omega L \) is the Inductive Reactance.
• \( X_C = \frac{1}{\omega C} \) is the Capacitive Reactance.

Step 1: {Analyze the condition for minimum impedance.}
To find the minimum value of \( Z \), we look at the term inside the square root. Since \( R^2 \) is a constant positive value, the impedance \( Z \) depends on the value of the reactance term \( (X_L - X_C)^2 \). Because this term is squared, its minimum possible value is \( 0 \). Therefore, \( Z \) is minimized when: \[ X_L - X_C = 0 \quad \Rightarrow \quad X_L = X_C \] This specific condition is called electrical resonance.

Step 2: {Substitute the expressions for reactance.}
Substituting the definitions of \( X_L \) and \( X_C \) into the resonance condition: \[ \omega L = \frac{1}{\omega C} \]

Step 3: {Solve for the angular frequency (\( \omega \)).}
Multiply both sides by \( \omega \): \[ \omega^2 L = \frac{1}{C} \] Divide both sides by \( L \): \[ \omega^2 = \frac{1}{LC} \] Taking the square root of both sides gives the resonant angular frequency: \[ \omega = \frac{1}{\sqrt{LC}} \] At this frequency, the impedance is purely resistive (\( Z = R \)).