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MCQ · Q2

Q.A resistor of 500 Ω\Omega and an inductance of 0.5 H are in series with an AC source which is given by V=1002sin⁡(1000t)V = 100\sqrt{2}\sin(1000t). The power factor of the combination is (A) 12\frac{1}{\sqrt{2}} (B) 13\frac{1}{\sqrt{3}} (C) 0.5 (D) 0.6

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The source is V=1002sin⁡(1000t)V=100\sqrt2\sin(1000t), so ω=1000\omega=1000 rad/s. With L=0.5L=0.5 H, the inductive reactance is XL=ωL=1000×0.5=500 ΩX_L=\omega L=1000\times0.5=500\,\Omega, which happens to equal R=500 ΩR=500\,\Omega. There is no capacitor in this circuit, so it is a series R-L combination with impedance Z=R2+XL2=5002+5002=5002 ΩZ=\sqrt{R^2+X_L^2}=\sqrt{500^2+500^2}=500\sqrt2\,\Omega. The power factor is cos⁡ϕ=RZ=5005002=12≈0.707\cos\phi=\dfrac{R}{Z}=\dfrac{500}{500\sqrt2}=\dfrac{1}{\sqrt2}\approx0.707. [!ANSWER] (A) 1/2≈0.7071/\sqrt2\approx0.707

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