Hello everyone, I have what appears to be a simple question, however i have a few questions regarding it..

My first question(well more of a check whether im doing it correctly) is to work out the mean value of the waveform V(t)=Acos(Wt)

Here is my solution:

_ T
V = 1/T ∫ Vdt where T= 2∏/W
0

_ 2∏/W
V = W/2∏ ∫ Acos(Wt)dt
0


_ 2∏/W
V = W/2∏ [AsinWT/W ]
0

_ 2∏/W
V = A /2∏ [sinWT ]
0

_
V = A /2∏ [sin2∏ -sin(0) ]

_
V = A /2∏

Does the final answer seem correct? I only question this because squaring the above, does not give me the correct answer of :
_
V = A^2 /2

Please tell me what I am doing wrong! Thanks

 

hmmm...the values on the integral have moved...For line 1, T and 0 are the integral limits, line 2 has limits 2∏/W and 0 and lines 3-4 have boundary conditions 2∏/W and 0. Sorry about that.

 

Here's a tip -- spend some time learning to use the LaTeX features of this site. They're pretty simple and make for easier-to-read equations. You're asking people for help, so make it easy for them to understand your question. I was willing to help, but because I can't read the equations easily, I just went to another post.

 

V = A /2∏ [sin2∏ -sin(0) ]
= A /2∏ [0 - 0]
= A /2∏ [0]
= 0

 

Thanks^^ @someonesdad, thankyou for the latex info, I really didnt realise how to do that. Im going to re-write the question again (even tho ErnieM helped with my issue) just to practice with it...

 

{tex}\omega = 2\cdot\pi\cdot\f {/tex}. This is just a test, please ignore...

 

\(\omega = 2\cdot\pi\cdot\f \).

 

i think if you wanted a short cut, to get the mean of a sinusoidal waveform you can multiply its peak by 0.606
handy if in a hurry, but only for sinusoidal waves

 

i think if you wanted a short cut, to get the mean of a sinusoidal waveform you can multiply its peak by 0.606
handy if in a hurry, but only for sinusoidal waves

You keep using that word.

I do not think it means what you thing it means.

 

my bad its actually 0.637 to get the mean.
What word am i confusing ErnieM? sinusoidal? if so i mean it in the context of a pure sine wave.

 

If by "mean" you mean "average" then the average value for any sine wave may be determined by multiplying a constant times either the peak value, the RMS value, or any instantaneous value, as the constant is always...

Zero.

There is no average voltage in a sine wave. The average value is the DC value. u-will-neva-no just showed a proof of that.

Also I believe .606 was a typo as you meant .707 which is the factor from peak value to RMS for a true sine wave.

Also, there is no parking in the red zone. The red zone is for loading and unloading only.

 

Well if we talking about sin wave.
For sure the average value for the whole period is equal 0.
But we can talk about average value for the half of the period .
Then the average value is equal

\( V_{AVG} = \frac{2}{\pi} * Vp = 0.636619772 * Vp\)

Where Vp - peak value

 

you can calculate a mean value, which is often, (maybe mistakenly granted) refered to as the average.
to calculate the mean you can multiply the peak by 0.637 (because the mean value of a sine wave is 63.7% of the peak). it cannot be used for say on the output of an SCR.
this site explains it a little better than i can i think!
http://www.electrowavecorp.com/images_and_links/power_measurements.htm

thats what i was trying to say Jony130,, i thought i was goin bonkers!