In the circuit shown, the value of L is 5H and the power factor is 0.8. It is also given that the voltage drop across capacitor is 2/5 times the voltage drop across the inductor. The impedance of the circuit (in ohm) is?

Power factor $\cos \mathrm{\varphi }=0.8=\frac{4}{5}$

$\therefore \mathrm{tan\varphi }=\sqrt{{\left(\frac{1}{\mathrm{cos\varphi }}\right)}^2-1}=\sqrt{{\left(\frac{5}{4}\right)}^2-1}=\frac{3}{4}$.

${\mathrm{X}}_{\mathrm{L}}=\mathrm{\omega L}=\left(314\times 5\right) \mathrm{\Omega }$.

${\mathrm{V}}_{\mathrm{C}}=\frac{2}{5}{\mathrm{V}}_{\mathrm{L}} \Rightarrow {\mathrm{X}}_{\mathrm{C}}=\frac{2}{5}{\mathrm{X}}_{\mathrm{L}}=\frac{2}{5}\times \left(314\times 5\right)=\left(314\times 2\right) \mathrm{\Omega }$.

From phasor diagram,

$\begin{array}{l}\mathrm{tan\varphi }=\frac{{\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}}{\mathrm{R}} \Rightarrow \frac{3}{4}=\frac{{\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}}{\mathrm{R}}\\ \Rightarrow {\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}=\frac{3}{4}\mathrm{R} \Rightarrow {\mathrm{X}}_{\mathrm{L}}\left(1-\frac{2}{5}\right)=\frac{3}{4}\mathrm{R}\end{array}$

$\Rightarrow \mathrm{R}=\frac{4}{5}{\mathrm{X}}_{\mathrm{L}}=\frac{4}{5}\times \left(314\times 5\right)=\left(314\times 4\right) \mathrm{\Omega }$.

Hence, impedance 

$\mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}=\sqrt{{\left(314\times 4\right)}^2+{\left(314\times 3\right)}^2}=314\times 5=1570 \mathrm{\Omega }$.