Answer to Question #311716 in Electric Circuits for Dreya

Question #311716

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

1
Expert's answer
2022-03-15T10:46:28-0400

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

a)Impedance (Z) of the coil

R=35ΩR=35\Omega

XL=2πfL=2π×50×0.18=56.6ΩX_L=2\pi fL=2\pi ×50×0.18=56.6\Omega

Z=R2+XL2=352+56.62=66.5ΩZ=\sqrt{R^2+X_L^2}=\sqrt{35^2+56.6^2}=66.5\Omega


b)Current (I) consumed by the coil

V=I×ZV=I×Z

I=VZ=10066.5=1.5A\therefore I=\frac{V}{Z}=\frac{100}{66.5}=1.5A


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

CosϕCos\phi =RZorSinϕ=XLZortanϕ=XLR=\frac{R}{Z}or Sin\phi =\frac{X_L}{Z} or tan\phi =\frac{X_L}{R}

Cosϕ=RZ=3566.5=0.5263\therefore Cos\phi =\frac{R}{Z}=\frac{35}{66.5}=0.5263

Cos1(0.5263)=58.2°Cos^{-1}(0.5263)=58.2°


d)Apparent power (S) consumed by the coil

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

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

S=V×I=100×1.5=150VAS=V×I=100×1.5=150VA



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