A resistor of `500 Omega` , an inductance of `0.5 H` are in series with an a.c. which is given by `V = 100 sqrt2 sin (1000 t)` . The power factor of the combination is

Options

A. `(1)/(sqrt2)`
B. `(1)/(sqrt3)`
C. 0.5
D. 0.6

Answer

A

Explanation

Here , `R = 500 Omega , L = 0.5 H`
Compare `V = 100 sqrt2` sin (1000 t) with V = `V_(0) sin omega t` ,
we get `omega = 1000`
The inductive reactance is `X_(L) = omega L = (1000) (0.5) = 500 Omega` Impedance of the RL circuit is `Z = sqrt(R_(2) + X_(L)^(2)) = sqrt((500 Omega)^(2) + (500 Omega)^(2)) = 500sqrt2 Omega`
Power factor , `cos phi (R)/(Z) = (500 Omega)/(500 sqrt2 Omega) = (1)/(sqrt2)`