Derive an expression for the impedance of an LCR circuit connected to an AC power supply. Draw phasor diagram.

Answer: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), with \(X_L = \omega L\), \(X_C = \dfrac{1}{\omega C}\).

A resistor \(R\), inductor \(L\) and capacitor \(C\) are connected in series to an a.c. source \(e = e_0\sin\omega t\). The same current \(i = i_0\sin(\omega t + \phi)\) flows through all of them.

• Voltage across R: \(V_R = i_0R\), in phase with current.
• Voltage across L: \(V_L = i_0X_L\), leading the current by \(\pi/2\).
• Voltage across C: \(V_C = i_0X_C\), lagging the current by \(\pi/2\).

Phasor diagram: [Diagram: current phasor \(i_0\) along the x-axis; \(V_R\) along the same direction; \(V_L\) vertically upward and \(V_C\) vertically downward; their resultant \((V_L - V_C)\) upward (taking \(V_L \gt V_C\)); the resultant of \(V_R\) and \((V_L - V_C)\) is \(e_0\), making angle \(\phi\) with the current phasor.]

Since \(V_L\) and \(V_C\) are opposite, their resultant is \(V_L - V_C\), which is perpendicular to \(V_R\). Hence

\(e_0^2 = V_R^2 + (V_L - V_C)^2 = i_0^2[R^2 + (X_L - X_C)^2]\)

\(\dfrac{e_0}{i_0} = \sqrt{R^2 + (X_L - X_C)^2}\).

The ratio \(e_0/i_0\) is the impedance: \(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + \left(\omega L - \dfrac{1}{\omega C}\right)^2}\).

The phase angle between e.m.f. and current: \(\tan\phi = \dfrac{X_L - X_C}{R}\).