Single Choice

A resistor of $$500 \Omega$$, an inductance of $$0.5 H$$ are in series with an a.c. which is given by $$V = 100\sqrt{2} \sin{\left(1000 t\right)}$$. The power factor of the combination is

A$$\dfrac{1}{\sqrt{2}}$$
Correct Answer
B$$\dfrac{1}{\sqrt{3}}$$
C$$0.5$$
D$$0.6$$

Solution

Given : $$L = 0.5 H$$ $$R = 500\Omega$$
On comparing source voltage with $$V = V_o \sin(wt)$$
We get $$w = 1000$$ $$s^{-1}$$
Inductive reactance $$X_L = wL = 1000\times 0.5 = 500\Omega$$
Power factor $$\cos\phi = \dfrac{R}{\sqrt{X_L^2 +R^2}}$$

$$\therefore$$ $$\cos\phi = \dfrac{500}{\sqrt{(500)^2 +(500)^2}}$$

Or $$\cos\phi = \dfrac{500}{500\sqrt{2}} = \dfrac{1}{\sqrt{2}}$$


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