Question

Derive the condition for resonance in the L-C-R series alternating current circuit. Find the expression for resonant frequency.

Hint:

Think of resonance as a "tug-of-war" between the inductor and the capacitor. The inductor wants to cause a phase lead, and the capacitor wants a phase lag. At resonance, their effects are equal and opposite (\(X_L = X_C\)), so they cancel each other out, leaving only the resistor to impede the current.

Answer 1

Step 1: Understanding the Concept:
Consider a series circuit containing an inductor (L), a capacitor (C), and a resistor (R) connected to an alternating voltage source \(V = V_m \sin(\omega t)\). The circuit offers opposition to the current flow, which is called impedance (\(Z\)). Resonance is a special condition in this circuit where the current becomes maximum.
Step 2: Key Formula or Approach:
The inductive reactance is \(X_L = \omega L\), which leads the current by a phase of \(\pi/2\).
The capacitive reactance is \(X_C = \frac{1}{\omega C}\), which lags behind the current by a phase of \(\pi/2\).
The total impedance (\(Z\)) of the series L-C-R circuit is given by the phasor diagram analysis:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] The peak current (\(I_m\)) in the circuit is given by Ohm's law for AC circuits:
\[ I_m = \frac{V_m}{Z} = \frac{V_m}{\sqrt{R^2 + (X_L - X_C)^2}} \] Step 3: Deriving the Condition for Resonance:
Electrical resonance occurs when the current in the circuit reaches its maximum possible value. From the equation for \(I_m\), it is clear that for a given \(V_m\) and R, the current \(I_m\) will be maximum when the impedance \(Z\) is minimum.
The impedance \(Z\) will be minimum when the term \((X_L - X_C)^2\) is minimum, which is zero.
\[ (X_L - X_C)^2 = 0 \] \[ X_L - X_C = 0 \] \[ X_L = X_C \] This is the condition for resonance in a series L-C-R circuit. At resonance, the inductive reactance is exactly equal to the capacitive reactance. The impedance becomes purely resistive, \(Z_{min} = R\), and the current is in phase with the voltage.
Step 4: Finding the Expression for Resonant Frequency:
Let the angular frequency at which resonance occurs be the resonant angular frequency, denoted by \(\omega_0\).
Using the resonance condition:
\[ X_L = X_C \] \[ \omega_0 L = \frac{1}{\omega_0 C} \] Rearranging the terms to solve for \(\omega_0\):
\[ \omega_0^2 = \frac{1}{LC} \] \[ \omega_0 = \frac{1}{\sqrt{LC}} \] This is the expression for the resonant angular frequency.
The ordinary frequency (\(f_0\)) is related to the angular frequency by \(\omega_0 = 2\pi f_0\).
\[ 2\pi f_0 = \frac{1}{\sqrt{LC}} \] \[ f_0 = \frac{1}{2\pi\sqrt{LC}} \] This is the expression for the resonant frequency. It is the natural frequency of oscillation of the circuit.