Question

A wound coil that has an inductance of 180mH and a resistance of 35 is connected to a 100V 50Hz supply calculate the impedance of the coil the current the power factor and the apparent power consumed

Answer 1

$R=35\Omega$ ,$L=180mH,V=100V,f=50Hz$

a)Impedance (Z) of the coil

$R=35\Omega$

$X_L=2\pi fL=2\pi ×50×0.18=56.6\Omega$

$Z=\sqrt{R^2+X_L^2}=\sqrt{35^2+56.6^2}=66.5\Omega$

b)Current (I) consumed by the coil

$V=I×Z$

$\therefore I=\frac{V}{Z}=\frac{100}{66.5}=1.5A$

c)The power factor and phase angle,$\Phi$

$Cos\phi$ $=\frac{R}{Z}or Sin\phi =\frac{X_L}{Z} or tan\phi =\frac{X_L}{R}$

$\therefore Cos\phi =\frac{R}{Z}=\frac{35}{66.5}=0.5263$

$Cos^{-1}(0.5263)=58.2°$

d)Apparent power (S) consumed by the coil

$P=V×I cos\phi =100×1.5cos(58.2°)=79W$

$Q=V×Isin\phi =100×1.5×sin(58.2°)=127.5VAr$

$S=V×I=100×1.5=150VA$