I searched through the data on this address http://www.allaboutcircuits.com/vol_2/chpt_2/4.html#02040.png but couldn't figure how they got a 53 whem they added 0 degrees and 90 degrees. Anyone know how?
Angular velocity is irrelevant, so long as it is the same for both voltages.I figured out that question and i now have yet another....
If
V1 = 80sin(ωt + 25°)
V2 = 10sin(ωt - 15°)
find the sum Vt=V1+V2 where the angular velocity (ω) is 400.
Where do you start?
2*Pi*f = w = 400now how would i calculate the frequency
P = 1/fperiod
You have to specify what the voltages given were.peak to peak voltage of V1, the rms voltage of V2
25 degrees - 15 degrees = 10 degreesthe phase difference between V1 and V1
(10/360)*Pand the time difference between the two.
I feel so guilty and ashamed for "helping" you so much. You will probably flunk out unless you can do problems like this yourself.Just set me up the equations and i'll have at it.
Ratch, do you see anything wrong with the first paragraph on this page:Hawkeye87,
Although the reference you cited talks about vectors, voltage is not a vector quantity. It has magnitide but not direction. Alternating voltages can be represented by phasors, which have magnitude and an angular relationship between them. Phasors also have some vector like properties, but not all of them. To add voltage phasors, split up each into real and orthogonal parts. Add up the real parts and the orthogonal parts. Then convert back to polar form if desired, or leave it like it is. scubasteve_911 touched on how to convert to polar form.
Ratch
No, I don't. But that is not the page that Hawkeye87 referenced.Ratch, do you see anything wrong with the first paragraph on this page: http://en.wikipedia.org/wiki/Phasor
I did help the OP, perhaps more that I should have. It is not a matter of semantics, it is about facts. Voltage is not a vector quantity.Let's help the O.P. with complex numbers, not argue about semantics.
I did not dwell on the mistake in the AAC, just a quick remark to get the OP on the right track. Which was helping him with his homework.Ratch: if you have a problem with the AAC website, make your comment in the appropriate forum - this one is for helping folk with homework.