Question

If power factor is $\frac{1}{2}$ in a series RL circuit. R = 100 $\Omega$, then L is (f = 50Hz)

• A. $\pi H$
• B. $\frac{\sqrt{3}}{\pi }H$
• C. $\frac{\pi }{\sqrt{3}}H$
• D. $2\pi H$

Answer 1

The correct option is B $\frac{\sqrt{3}}{\pi }H$

given $R=100\Omega$

$f=50Hz$

the power factor of R-L series circuit$=\frac{R}{\sqrt{R^2+{X_L}^2}}$

$\frac{1}{2}=\frac{R}{\sqrt{R^2+\frac{{\omega }^2}{L^2}}}$

$\frac{1}{2}=\frac{1}{\sqrt{1+\frac{{\omega }^2{L}^2}{R^2}}}=\frac{1}{\sqrt{1+\frac{{\left(50\right)}^2\times L^2}{{100}^2}}}$

after squaring and solving we get$L=\frac{\sqrt{3}}{\pi }H$