Question

The measured power, current, and voltage delivered to a load are 100W, 1A, and 120V respectively.?
How would I find the reactive current? http://static.panoramio.com/photos/medium/94840337.jpg

Answer 1

would I get the answer by doing sin 34 degrees

Answer 2

Hi! I'm just learning this on Wikipedia right now. But it seems that the real power is \(Q=V_\text{rms}\ I_\text{rms}\ \cos(\phi)\) where \(\phi\) is the phase angle.

So \(\phi=\cos^{-1}\left(\dfrac{Q}{V_\text{rms}\ I_\text{rms}}\right)\approx34^\circ\)

So I see how you got that. And then the reactive component of current seems like it would be \(I_\text{rms}\ \sin(\phi)\) because the sine is the reactive component. I believe \(I_X\) is the reactive current. After all, I know \(X\) is reactance.

Answer 3

Well done @theEric, and since the total (or resultant) current is leading the resistance current it is inductive reactance\[X _{L}\]

Answer 4

Thanks, @radar ! :)

Answer 5

So i'm right?

Answer 6

You didn't show how you got the 34 degrees, which is a step in the right direction in getting the solution.

The apparent power VA is E X I, and in this case 120 V X 1 Amp or 120 VA.
The actual or true power was given as 100 Watts. You have learned the the real power is equal to the apparent power times the cosine of the phase difference in the voltage vs current. With the information you now have you can calculate this phase angle.

100 = (120) Cos phase angle
Cos phase angle = 100/120 = .83333
Now to find the actual phase angle take the inverse cos of .83333 = 33.557 degree.
You now have the phase angle. Please refer to the triangle in your link.

You have the total current provided as a value of 1 amp, and from the triangle the reactive curren (which is what the problem is requesting) would be:\[I _{X}=(Sin 33.557) X 1 A = 0.55277 A\]
Is this what you calculated in your solution? Note the units is Amperes

Answer 7

Yeah I already showed how i got the 34 degrees. I just wanted to make sure I was right. Thanks for your help again!