Question

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

Answer 1

Answer:

Apparent power is 1.413Volt-Amps.

Step-by-step 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=$\frac{E_{T}^{2}}{R}=\frac{208^{2}}{33000}=1.311watts$

Reactive power=$\frac{E_{T}^{2}}{X_{L}}=\frac{208^{2}}{82000}=0.528VARs$

Hence apparent power=$\sqrt{1.311^{2}+0.528^{2}} =1.413VA$