Question

Using a phasor diagram, derive an expression for the impedance of a series LCR circuit connected to an ac source. What is meant by resonance in this circuit? Obtain expression for the resonant frequency of this circuit.

Hint:

Draw V_R along the current, V_L up and V_C down. Add by Pythagoras for Z. Resonance is when X_L = X_C.

Answer 1

Step 1: Setting up the circuit:
A resistor \(R\), an inductor \(L\) and a capacitor \(C\) are joined in series to the source \(v = v_m \sin \omega t\).
The same current \(i = i_m \sin(\omega t + \phi)\) flows through all three parts. So we draw the current phasor as the reference.

Step 2: Phasors of the three voltages:
Resistor: \(V_R = i_m R\). It is in phase with the current, so it is drawn along the current phasor.
Inductor: \(V_L = i_m X_L\) with \(X_L = \omega L\). It leads the current by \(90^{\circ}\), so it is drawn perpendicular to the current phasor, upward.
Capacitor: \(V_C = i_m X_C\) with \(X_C = \dfrac{1}{\omega C}\). It lags the current by \(90^{\circ}\), so it is drawn perpendicular to the current phasor, downward.
Diagram in words: draw a horizontal arrow for \(i_m\) and \(V_R\). Draw \(V_L\) up from its tip and \(V_C\) down. Take \(V_L > V_C\). The resultant \(v_m\) is the hypotenuse of the right triangle with sides \(V_R\) and \(V_L - V_C\). It makes an angle \(\phi\) with the current phasor.

Step 3: Expression for impedance:
The source voltage phasor is the vector sum of the three. By Pythagoras:
\[ v_m^2 = V_R^2 + (V_L - V_C)^2 = i_m^2\,[\,R^2 + (X_L - X_C)^2\,] \]
\[ v_m = i_m \sqrt{R^2 + (X_L - X_C)^2} \]
The impedance is \(Z = \dfrac{v_m}{i_m}\), so:
\[ Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \]
The phase angle between voltage and current is given by \(\tan\phi = \dfrac{X_L - X_C}{R}\).

Step 4: Meaning of resonance:
Resonance is the condition in which the inductive reactance equals the capacitive reactance, \(X_L = X_C\). At this point \(V_L\) and \(V_C\) cancel.
Then \(Z = R\) is the minimum value, the current \(i_m = v_m/R\) is the maximum, and the current is in phase with the voltage.

Step 5: Resonant frequency:
Put \(X_L = X_C\) at the resonant angular frequency \(\omega_0\):
\[ \omega_0 L = \frac{1}{\omega_0 C} \quad \Rightarrow \quad \omega_0 = \frac{1}{\sqrt{LC}} \]
\[ \nu_0 = \frac{\omega_0}{2\pi} = \frac{1}{2\pi\sqrt{LC}} \]

Final Answer:
Impedance is \(Z = \sqrt{R^2 + (\omega L - 1/\omega C)^2}\). At resonance \(X_L = X_C\), \(Z = R\), and the resonant frequency is \(\nu_0 = \dfrac{1}{2\pi\sqrt{LC}}\). \[ \boxed{Z = \sqrt{R^2 + (\omega L - 1/\omega C)^2}, \quad \nu_0 = \frac{1}{2\pi\sqrt{LC}}} \]