A series LCR circuit is formed using a 200mH inductor, a 500μF capacitor and a 10Ω resistor. This circuit is connected across a 100 V variable frequency AC source. Find out the particular frequency at which the power factor of the circuit is unity.

Here

$\begin{array}{l}\mathrm{L}=200\mathrm{mH}=200\times {10}^{-3}\mathrm{H}=0.2\mathrm{H}\\ \mathrm{C}=500\mathrm{\mu F}=500\times {10}^{-6}=5\\ \times {10}^{-4}\mathrm{F}\\ \mathrm{R}=10\mathrm{\Omega }\text{ and }{\mathrm{V}}_{\mathrm{rms}}=100\mathrm{V}\\ \text{ Power factor }=\cos \mathrm{\varphi }=\frac{\mathrm{R}}{\mathrm{Z}}=1\text{ (Given) }\\ \therefore \mathrm{Z}=\mathrm{R}\Rightarrow \sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}=\mathrm{R}\\ {\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2={\mathrm{R}}^2\Rightarrow {\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\\ =0\end{array}$

or X_L = X_C (this is resonance condition)

$\begin{array}{l}2\mathrm{\pi vL}=\frac{1}{2\mathrm{\pi vC}}\Rightarrow 4{\mathrm{\pi }}^2{\mathrm{v}}^2\mathrm{LC}=1\\ \therefore \mathrm{v}=\frac{1}{2\mathrm{\pi }\sqrt{\mathrm{LC}}}=\frac{1}{2\times 3.14\times \sqrt{0.2\times 5\times {10}^{-4}}}\\ =\frac{1}{2\times 3.14\times {10}^{-2}}=\frac{100}{6.28}=15.9\mathrm{Hz}\end{array}$