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  • A resistor of 500,an inductance of 0.5 H are in series with an ac which is given by V = sin (1000t). The power fac

A resistor of 500\Omega,an inductance of 0.5 H are in series with an ac which is given by V = sin (1000t). The power factor of the combination is:

Option: 1

\frac{1}{\sqrt{2}}


Option: 2

\frac{1}{\sqrt{3}}


Option: 3

0.5


Option: 4

0.6


Answers (1)

best_answer

Here, \mathrm{\mathrm{R}=500 \Omega, \mathrm{L}=0.5 \mathrm{H} }
Compare \mathrm{\mathrm{V}=100 \sqrt{2} \sin (1000 \mathrm{t}) with \mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}, we \, \, get \, \, \omega=1000 }
The inductive reactance is \mathrm{\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=(1000)(0.5)=500 \Omega }
Impedance of the RL circuit is 
\mathrm{\mathrm{Z}=\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2}=\sqrt{(500 \Omega)^2+(500 \Omega)^2}=500 \sqrt{2} \Omega }
Power factor,
 \mathrm{\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{500 \Omega}{500 \sqrt{2} \Omega}=\frac{1}{\sqrt{2}} }

Posted by

manish painkra

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