Question

A series combination of L, C and R is connected to an a.c. source. Using a phasor diagram, derive an expression for the impedance of the circuit and phase difference between V and I.

Hint:

When $X_L > X_C$, the circuit is predominantly inductive, and voltage leads current.
When $X_C > X_L$, the circuit is predominantly capacitive, and voltage lags behind current.
At resonance ($X_L = X_C$), $Z = R$ and $\phi = 0^\circ$, meaning voltage and current are in phase.

Answer 1

Concept:
• In a series LCR circuit, an inductor $L$, capacitor $C$, and resistor $R$ are connected in series across an alternating voltage source $V = V_0 \sin(\omega t)$.

• The current $I$ is common to all three components at any instant.

• Voltage across resistor $V_R = I R$ is in phase with current $I$.

• Voltage across inductor $V_L = I X_L$ leads current $I$ by $\frac{\pi}{2}$ radians, where $X_L = \omega L$.

• Voltage across capacitor $V_C = I X_C$ lags behind current $I$ by $\frac{\pi}{2}$ radians, where $X_C = \frac{1}{\omega C}$.

Step 1: Phasor Diagram Construction
Let current phasor $\vec{I}$ be drawn along the positive $x$-axis.
The potential difference across resistor $\vec{V}_R$ is along the $x$-axis (in phase with $\vec{I}$).
The potential difference across inductor $\vec{V}_L$ is along the positive $y$-axis (leading $\vec{I}$ by $90^\circ$).
The potential difference across capacitor $\vec{V}_C$ is along the negative $y$-axis (lagging $\vec{I}$ by $90^\circ$).
Assuming $V_L > V_C$, the resultant vector along the $y$-axis is $(V_L - V_C)$ pointing along the positive $y$-axis.

Step 2: Derivation of Impedance Expression
Using vector addition for phasors, the total applied voltage $V$ is given by the hypotenuse of the right-angled triangle formed by $V_R$ and $(V_L - V_C)$:
\[ V^2 = V_R^2 + (V_L - V_C)^2 \]
Substitute $V_R = I R$, $V_L = I X_L$, and $V_C = I X_C$:
\[ V^2 = (I R)^2 + (I X_L - I X_C)^2 \]
\[ V^2 = I^2 \left[ R^2 + (X_L - X_C)^2 \right] \]
Taking the square root on both sides:
\[ V = I \sqrt{R^2 + (X_L - X_C)^2} \]
The total effective opposition offered by the series LCR circuit to alternating current is called impedance $Z$:
\[ Z = \frac{V}{I} = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$:
\[ Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \]

Step 3: Derivation of Phase Difference
From the phasor diagram, if $\phi$ is the phase angle between the total supply voltage $V$ and current $I$:
\[ \tan \phi = \frac{V_L - V_C}{V_R} = \frac{I X_L - I X_C}{I R} = \frac{X_L - X_C}{R} \]
\[ \phi = \tan^{-1} \left( \frac{\omega L - \frac{1}{\omega C}}{R} \right) \]

Step 4: Conclusion
The impedance of the series LCR circuit is $Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$ and the phase difference between voltage and current is $\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$.