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A resistor of 500 Omega and an inductanc...

A resistor of `500 Omega` and an inductance of 0.5 H are in series with an ac source which is given by `V=100sqrt(2)sin(1000t)`. The power factor of the combination is

A

`(1)/(sqrt2)`

B

`(1)/(sqrt3)`

C

0.5

D

0.6

Text Solution

Verified by Experts

The correct Answer is:
A
Here , ` R = 500 Omega, L = 0.5H`
Compare `V = 100 sqrt2 sin (100t)` with `V = V_0 sin omega t `
we get , `W = 1000 rad s^(-1)`
The inductive reactance is `X_L = omega L = (1000)(0.5) = 500 Omega`
Impedance of the LR circuit is
` Z = sqrt(R^2 + X_L^2) = sqrt((500 Omega)^2 + (500 Omega)^2) = 500 sqrt2 Omega`
Power factor, `cos phi = R/Z = (500 Omega)/(500 sqrt2 Omega) = (1)/(sqrt2)`
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