Question

In a parallel circuit, ET = 208 V, R = 33 kΩ, and XL = 82 kΩ. What is the apparent power?

Answer 1

Apparent power is 1.413Volt-Amps.

Explanation:

We are given that in a parallel circuit,
`E_(T)=208V`,R=33kΩ and
`X_(L)`=82kΩ.

And we are asked to find apparent power.

Apparent power is the combination of true power and reactive power.

In other words, we can say that in a power triangle apparent power is the hypotenuse where as true power and reactive powers are sides.

Let us find true power , reactive power first and the using Pythagorean theorem we can find apparent power.

True power=
`(E_(T)^(2))/(R)=(208^(2))/(33000)=1.311watts`

Reactive power=
`(E_(T)^(2))/(X_(L))=(208^(2))/(82000)=0.528VARs`

Hence apparent power=
`\sqrt{1.311^(2)+0.528^(2)} =1.413VA`