Obtain an expression for average power dissipated in a series LCR circuit.

Answer: \(P_{av} = e_{rms}\,i_{rms}\cos\phi\)

Let an alternating e.m.f. \(e = e_0\sin\omega t\) be applied to a series LCR circuit. The current is \(i = i_0\sin(\omega t - \phi)\), where \(i_0 = e_0/Z\), \(Z = \sqrt{R^2 + (X_L - X_C)^2}\) and \(\tan\phi = \dfrac{X_L - X_C}{R}\).

Instantaneous power:

Average power over one complete cycle (period \(T\)): the average of \(\sin^2\omega t\) is \(\frac{1}{2}\) and the average of \(\sin\omega t\cos\omega t = \frac{1}{2}\sin 2\omega t\) is 0. Hence

Here \(\cos\phi = \dfrac{R}{Z}\) is the power factor. Since \(e_{rms} = i_{rms}Z\), also \(P_{av} = i_{rms}^2 R\): power is dissipated only in the resistance.

• Purely resistive circuit or resonance: \(\phi = 0\), \(\cos\phi = 1\), \(P_{av} = e_{rms}i_{rms}\) (maximum).
• Purely inductive or capacitive circuit: \(\phi = 90^\circ\), \(\cos\phi = 0\), \(P_{av} = 0\) (wattless current).