Question

What is the correct condition for an LCR circuit to be at resonance?

Answer 1

An LCR (inductor-capacitor-resistor) circuit is said to be at resonance when the inductive reactance and capacitive reactance in the circuit cancel out each other, leaving only the resistance in the circuit. This results in maximum current flow through the circuit and maximum energy transfer between the inductor and the capacitor.

The condition for an LCR circuit to be at resonance is given by the resonance frequency formula:

Resonance Frequency Formula:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]

where:

• f is the frequency of the AC source
• L is the inductance of the inductor in the circuit, measured in henries (H)
• C is the capacitance of the capacitor in the circuit, measured in farads (F)
• \(\pi\) is the mathematical constant pi (approximately equal to 3.14)

At resonance, the inductive reactance (\(X_L\)) and capacitive reactance (\(X_C\)) in the circuit are equal in magnitude but opposite in sign, i.e., \(X_L = -X_C\). The impedance of the circuit (\(Z\)) is given by:

\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

where:

• R is the resistance in the circuit, measured in ohms (Ω)

At resonance, since \(X_L = -X_C\), the term \((X_L - X_C)\) in the above equation becomes zero. Therefore, the impedance of the circuit simplifies to:

\[ Z = R \]

This means that the circuit behaves like a pure resistive circuit, with maximum current flow through it and maximum power transfer between the inductor and the capacitor.

Conclusion: The condition for an LCR circuit to be at resonance is when the frequency of the AC source is equal to the resonance frequency given by the formula:

\[ f = \frac{1}{2\pi\sqrt{LC}} \]

and the inductive reactance and capacitive reactance in the circuit are equal in magnitude but opposite in sign.