Question

The series combination of resistance R and inductance L is connected to an alternating source of e. m. f. e = 311 sin (100 πt). If the value of wattless current is 0.5 A and the impedance of the circuit is 311 Ω, the power factor will be - ​

Answer 1

The power factor is cos (Φ) = 1 / √2

Explanation:

Power factor can found using the equation = cos (Φ)    

Φ is the difference of phases.

While (Φ) = I' / I (rms) Where I' according to the given question runs with watt-less current.

So, Eo / Io = I (impedence) and E = Eo sin (ωt)

The power factor formula is Sin (Φ)  = 1 /2 ÷ 1 / √2, Φ = π / 4

Hence we get, Power factor = cos (Φ) = 1 / √2