Question

A coil of inductance 80 mH is joined in series with a resistance of $60\Omega$. An A.C EMF of 220 V & $\frac{500}{\pi }Hz$ is applied to it. Wattless current in the circuit will be (Write upto two digits after the decimal point)

Answer 1

$X_L=\omega L=2\pi fL=2\pi \frac{500}{\pi }80\times {10}^{-3}=80\Omega$
$R=60\Omega$
So $Z=\sqrt{X_L^2+R^2}=100\Omega \Rightarrow i=\frac{\in }{z}=\frac{220}{100}=2.2A$

$\tan \varphi =\frac{80}{60}\Rightarrow \varphi ={53}^{\circ }$
Wattless current $i_{WL}=i\sin 53=2.2\times \frac{4}{5}=1.76$