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:

For a series LCR circuit, always remember the two most important formulas: \[ Z=\sqrt{R^2+(X_L-X_C)^2} \] and \[ \tan\phi=\frac{X_L-X_C}{R} \] These formulas directly determine the impedance and phase relationship between voltage and current.

Answer 1

Concept: A series LCR circuit consists of a resistor (\(R\)), an inductor (\(L\)) and a capacitor (\(C\)) connected in series to an alternating voltage source. In an AC circuit, the voltages across the resistor, inductor and capacitor are generally not in the same phase. Therefore, simple algebraic addition of voltages is not possible. Instead, vector addition using a phasor diagram is employed. The opposition offered by the LCR circuit to the flow of alternating current is called its impedance. \[ Z=\frac{V}{I} \] where
• \(Z\) = impedance,
• \(V\) = rms voltage,
• \(I\) = rms current.

Step 1: Voltages across individual circuit elements Let the alternating current flowing through the series LCR circuit be \[ I=I_0\sin\omega t \] Since the circuit elements are connected in series, the same current flows through all of them. Voltage across resistor For a resistor, \[ V_R=IR \] The voltage across the resistor is in phase with the current. Voltage across inductor For an inductor, \[ V_L=IX_L \] where \[ X_L=\omega L \] is the inductive reactance. The voltage across the inductor leads the current by \(90^\circ\). Voltage across capacitor For a capacitor, \[ V_C=IX_C \] where \[ X_C=\frac{1}{\omega C} \] is the capacitive reactance. The voltage across the capacitor lags the current by \(90^\circ\).

Step 2: Construction of phasor diagram Taking current \(I\) as the reference phasor:
• \(V_R\) is drawn along the direction of current.
• \(V_L\) is drawn vertically upward because it leads current by \(90^\circ\).
• \(V_C\) is drawn vertically downward because it lags current by \(90^\circ\). The net reactive voltage is \[ V_L-V_C \] Thus, the resultant voltage \(V\) is obtained by vector addition of \(V_R\) and \((V_L-V_C)\). Phasor Diagram

The resultant voltage \(V\) is the diagonal of the right-angled triangle formed by \(V_R\) and \((V_L-V_C)\).

Step 3: Determine the resultant voltage Using Pythagoras theorem, \[ V^2=V_R^2+(V_L-V_C)^2 \] Substituting \[ V_R=IR,\qquad V_L=IX_L,\qquad V_C=IX_C \] we get \[ V^2=(IR)^2+\left(IX_L-IX_C\right)^2 \] \[ V^2=I^2\left[R^2+(X_L-X_C)^2\right] \] Taking square root on both sides, \[ V=I\sqrt{R^2+(X_L-X_C)^2} \]

Step 4: Expression for impedance Since \[ Z=\frac{V}{I} \] therefore, \[ \boxed{ Z=\sqrt{R^2+(X_L-X_C)^2} } \] Substituting \[ X_L=\omega L, \qquad X_C=\frac{1}{\omega C} \] we obtain \[ \boxed{ Z= \sqrt{ R^2+ \left( \omega L-\frac{1}{\omega C} \right)^2 } } \] This is the required expression for the impedance of a series LCR circuit.

Step 5: Determine the phase difference Let \(\phi\) be the phase difference between the applied voltage and current. From the phasor triangle, \[ \tan\phi = \frac{V_L-V_C}{V_R} \] Substituting \[ V_L=IX_L, \qquad V_C=IX_C, \qquad V_R=IR \] we get \[ \tan\phi = \frac{IX_L-IX_C}{IR} \] Cancelling \(I\), \[ \tan\phi = \frac{X_L-X_C}{R} \] Hence, \[ \boxed{ \tan\phi= \frac{X_L-X_C}{R} } \] or \[ \boxed{ \phi= \tan^{-1} \left( \frac{X_L-X_C}{R} \right) } \] Interpretation of the phase angle
• If \(X_L>X_C\), then \(\phi>0\) and the circuit behaves inductively. Current lags voltage.
• If \(X_C>X_L\), then \(\phi<0\) and the circuit behaves capacitively. Current leads voltage.
• If \(X_L=X_C\), then \(\phi=0\) and voltage and current are in phase. Final Result: The impedance of a series LCR circuit is \[ \boxed{ Z= \sqrt{ R^2+ (X_L-X_C)^2 } } \] and the phase difference between voltage and current is \[ \boxed{ \tan\phi= \frac{X_L-X_C}{R} } \] or \[ \boxed{ \phi= \tan^{-1} \left( \frac{X_L-X_C}{R} \right) } \]