Derive an expression for the impedance of a series LCR circuit connected to an AC power supply.

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

Let a resistor \(R\), an inductor \(L\) and a capacitor \(C\) be 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 three. Let the peak voltages across them be \(V_R\), \(V_L\), \(V_C\):

• \(V_R = i_0R\), in phase with the current;
• \(V_L = i_0X_L\), leading the current by \(\pi/2\), with \(X_L = \omega L\);
• \(V_C = i_0X_C\), lagging the current by \(\pi/2\), with \(X_C = 1/\omega C\).

[Figure: phasor diagram — current phasor \(i_0\) and \(V_R\) along the x-axis; \(V_L\) along +y, \(V_C\) along −y; the resultant of \((V_L - V_C)\) and \(V_R\) is \(e_0\), making angle φ with the current.]

\(V_L\) and \(V_C\) are in opposite directions, so their resultant is \(V_L - V_C\) (taking \(V_L \gt V_C\)), perpendicular to \(V_R\). Hence the applied peak emf is

The impedance is \(Z = e_0/i_0\):

The phase angle between emf and current is given by \(\tan\phi = \dfrac{X_L - X_C}{R}\).