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

Options

A. `(1)/( sqrt(2))`
B. `(1) /( sqrt(3))`
C. `0.5`
D. `0.6`

Answer

A

Explanation

`R=500 Omega `
`X_(L )= Omega xx L = 1000 xx 0.5 = 500`
` cos phi = ( R )/(sqrt(R^(2)+X_(L )^(2)))`