Question

Calculate the Power Factor for a series LCR circuit with a source of frequency f = 50. If the connected across $L=\left(\pi /2\right)H,C=\left(\pi \right)\mu F$ and $R=500\Omega$. Take ${\pi }^2=10$

• A. 1
• B. 0
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$

$\omega =2\pi f=100\pi X_L=\omega L=100\pi \times \frac{\pi }{2}=500\Omega ({\therefore \pi }^2=10)X_C=\frac{1}{\omega C}=\frac{1}{100\pi \times \pi \times {10}^{-6}}=1000\Omega Z=\sqrt{R^2+{(X_L-X_C)}^2}=\sqrt{{500}^2+{(500-1000)}^2}=500\sqrt{2}\Omega PowerFactor=\cos \varphi =\frac{R}{Z}=\frac{500}{500\sqrt{2}}=\frac{1}{\sqrt{2}}$

Answer will be option (D).