Question

In a parallel circuit, ET = 208 V, R = 39 kΩ, and XL = 24 kΩ. What is the power factor?

Answer 1

Power factor is 0.8517 or 85.17%

Explanation:

We are given that in a parallel circuit,
`E_(T)=208V`, R=39kΩ and
`X_(L)`=24kΩ.

That means the circuit has a resistor element, inductor element in parallel with 208V applied across the terminal.

Let us find the current through resistor,

`I_(R) =\frac{E_(T)} {R}= (208)/(39) =(16)/(3)=5.33` milli amps

And current through inductor,
`I_(L) =\frac{E_(T)} {X_(L)} =(208)/(24)=(26)/(3)=8.667` milli amps.

Hence total current,
`I_(T)= \sqrt{(I_(R))^2+(I_(T))^(2)} =\sqrt{5.33^(2)+8.667^(2)}=10.176`

Now power factor=
`(I_(R))/(I_(T))=(5.33)/(10.176)=0.8517`

Hence power factor is 85.17% or 0.8517