Question

A resistance of $300\Omega$ and an inductance of $1/\pi$ Henry are connected in series to an ac voltage of 20 volts and 200 Hz frequency. The phase angle between the voltage and current is $ta{n}^{-1}\left(x/y\right)$. Find $xy$.

Answer 1

If phase angle between voltage and current is $\varphi$ then from phasor diagram,

$tan\varphi =\frac{V_L}{V_R}=\frac{i{X}_L}{i_R}=\frac{X_L}{R}$

Where, inductive reactance,

$X_L=\omega L=2\pi fL$

$tan\varphi =\frac{2\pi fL}{R}=\frac{2\pi \times 200Hz\times \frac{1}{\pi }H}{300\Omega }=\frac{4}{3}$

$\therefore \varphi =ta{n}^{-1}\frac{4}{3}$

Comparing with given value,

$\Rightarrow \varphi =ta{n}^{-1}\frac{x}{y}=ta{n}^{-1}\frac{4}{3}$

$\Rightarrow x=4,y=3$

$\Rightarrow xy=12$

Hence, value of xy is 12.