Question

(a) An ac voltage of frequency \(\omega\) is applied across a series LCR circuit. Draw the phasor diagram and obtain the impedance of the circuit.

(b) Discuss 'resonance' in a series LCR circuit and write the expression for resonant frequency.

Hint:

Add the phasors V_R, V_L and V_C like vectors; at resonance X_L = X_C.

Answer 1

Step 1: Voltage across each element:
Let the source be \(v = v_m \sin \omega t\). In a series circuit the same current \(i\) flows through R, L and C, so we compare the three voltages with this current.
Across the resistor, \(V_R = IR\), and it is in phase with the current.
Across the inductor, \(V_L = I X_L\) with \(X_L = \omega L\). It leads the current by \(90^{\circ}\).
Across the capacitor, \(V_C = I X_C\) with \(X_C = \dfrac{1}{\omega C}\). It lags the current by \(90^{\circ}\).

Step 2: Phasor diagram (described):
Draw the current phasor \(\vec{I}\) along the horizontal axis as the reference. Draw \(\vec{V}_R\) along \(\vec{I}\).
Draw \(\vec{V}_L\) vertically upward and \(\vec{V}_C\) vertically downward, both starting from the same origin.
Take \(V_L > V_C\). The net reactive phasor is \(\vec{V}_L - \vec{V}_C\), pointing upward. The source phasor \(\vec{V}\) is the resultant of \(\vec{V}_R\) and \((\vec{V}_L - \vec{V}_C)\), and it makes an angle \(\phi\) with \(\vec{I}\).

Step 3: Impedance from the diagram:
The phasors form a right-angled triangle, so by Pythagoras:
\[ V^2 = V_R^2 + (V_L - V_C)^2 = I^2\left[R^2 + (X_L - X_C)^2\right] \]
Impedance is \(Z = V/I\), so
\[ Z = \sqrt{R^2 + \left(\omega L - \dfrac{1}{\omega C}\right)^2} \]
The phase angle follows from \(\tan\phi = \dfrac{V_L - V_C}{V_R} = \dfrac{X_L - X_C}{R}\).

Step 4: Resonance:
When the source frequency is changed, \(X_L\) rises and \(X_C\) falls. At one frequency \(X_L = X_C\), and the circuit is said to be in resonance.
Then \(\omega_0 L = \dfrac{1}{\omega_0 C}\), which gives
\[ \omega_0 = \dfrac{1}{\sqrt{LC}}, \qquad f_0 = \dfrac{1}{2\pi\sqrt{LC}} \]
At resonance the impedance is minimum, \(Z = R\). The current is maximum, \(I_{max} = V/R\), and it is in phase with the voltage (\(\phi = 0\)). The circuit behaves as a pure resistor, and \(V_L\) and \(V_C\) are equal and opposite.
Below \(f_0\) the circuit is capacitive (\(X_C > X_L\)). Above \(f_0\) it is inductive (\(X_L > X_C\)).

Final Answer:
The impedance is \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), and the resonant frequency is \(f_0 = \dfrac{1}{2\pi\sqrt{LC}}\), where \(Z = R\) is minimum and the current is maximum.\[ \boxed{Z = \sqrt{R^2 + (X_L - X_C)^2}, \quad f_0 = \dfrac{1}{2\pi\sqrt{LC}}} \]