In the LCR series AC circuit shown, the r.m.s. readings of voltmeters are V₁ = 150V, V₂ = 50V and the r.m.s. voltage of the source is 130 V. Then

$$\begin{gathered} {{\mathrm{V}}_{\mathrm{R}}}^2+{{\mathrm{V}}_{\mathrm{L}}}^2={{\mathrm{V}}_1}^2={150}^2 \cdots \left(\mathrm{i}\right) \\ \left|{\mathrm{V}}_{\mathrm{L}}-{\mathrm{V}}_{\mathrm{C}}\right|={\mathrm{V}}_2=50 \cdots \left(\mathrm{ii}\right) \\ {{\mathrm{V}}_{\mathrm{R}}}^2+{\left({\mathrm{V}}_{\mathrm{L}}-{\mathrm{V}}_{\mathrm{C}}\right)}^2={\mathrm{V}}^2={130}^2 \cdots \left(\mathrm{iii}\right) \end{gathered}$$

Using (ii) and (iii), ${{\mathrm{V}}_{\mathrm{R}}}^2+{50}^2={130}^2 \Rightarrow {\mathrm{V}}_{\mathrm{R}}=\sqrt{{130}^2-{50}^2}=120 \mathrm{V}$.

From (i), ${\mathrm{V}}_{\mathrm{L}}=\sqrt{{150}^2-{120}^2}=90 \mathrm{V}$.

Using (ii), ${\mathrm{V}}_{\mathrm{C}}={\mathrm{V}}_{\mathrm{L}}\pm 50=90\pm 50=140 \mathrm{V} \text{or} 40 \mathrm{V}$.

Power factor, $\mathrm{cos\varphi }=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{{\mathrm{V}}_{\mathrm{R}}}{\mathrm{V}}=\frac{120}{130}=\frac{12}{13}$.