In an L-R-C series circuit, R = 150 Ω, L = 0.750 H, and C = 0.0180 mF. The source has voltage amplitude V = 150 V and a frequency equal to the resonance frequency of the circuit. (a) What is the power factor? (b) What is the average power delivered by the source? (c) The capacitor is replaced by one with C = 0.0360 mF and the source frequency is adjusted to the new resonance value. Then what is the average power delivered by the source?

The angle between the voltage and current phasors is given by equation

$\mathbf{tan\varphi }\mathbf{=}\left(\frac{X_L-X_C}{R}\right)$, the average power supplied by resistor only is ${\mathbf{P}}_{\mathbf{av}}\mathbf{=}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{2}\mathbf{R}}$.

Since it is given that inductive resistance and capacitive reactance are same,

$$\begin{gathered} X_L=X_C \\ \tan \varphi =\left(\frac{X_L-X_C}{R}\right) \\ =\left(\frac{X_C-X_C}{R}\right) \\ =0 \\ \varphi =0° \end{gathered}$$

Hence, the power factor $\cos \varphi =1$.

The average rate here represents the average power supplied to the resistor only, so

$$\begin{gathered} {\mathrm{P}}_{\mathrm{av}}=\frac{{\mathrm{V}}^2}{2\mathrm{R}} \\ =\frac{{\left(150\mathrm{V}\right)}^2}{2\times 150\mathrm{\Omega }} \\ =75\mathrm{W} \end{gathered}$$

The new capacitance is $\mathrm{C}=0.0360\mathrm{\mu F}$, but in resonance case, the inductive reactance is the same as capacitive reactance. The voltage in the inductor and the capacitor cancel each other and the amplitude voltage is kept the same. Also, the impedance is equal to the resistance.

As the average power depends on R and V which they are the same, therefore, the average power will be the same as in part(b)

Hence, the average power will be the same as in part (b) i.e. 75W.

Hence, (a) Power factor is 1.00, (b) average power delivered by the source is 75W,(c)average power delivered by the source is 75W.