# Phasors, square root of two times sine = ?

Discussion in 'Homework Help' started by kvnsmith.at.uw, May 23, 2012.

1. ### kvnsmith.at.uw Thread Starter New Member

May 16, 2012
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0
I've been watching videos from this guy, http://www.youtube.com/watch?v=ikeIWcBoXUI , and I find them fairly useful as far as procedure, but in our classes we are told to convert everything to cos which he does not do. Further, his equations are are √2 times sin, but he leaves off the √2 during calculations. More specifically, 100√2 sin(bla) gives what he calls the RMS voltage of 100, ignoring the root and ignoring the fact that it's a sin. I believe that it works, I just don't know what he's doing.

2. ### WBahn Moderator

Mar 31, 2012
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4,917
Have they taught you what the meaning of the RMS equivalent voltage of a signal is?

Real quickly, given a voltage signal v(t) from an ideal voltage source (i.e., zero output impedance), there is a certain amount of power that will be delivered to a purely resistive load, R. While the instaneous power delivered will vary as v(t) varies, you can calculate the average power delivered over a sufficiently long period of time (if the voltage signal is periodic, then the average power of one period is what you are looking for.

So, do the calculation and you will have Pavg = f(v(t), R) or, in words, the average power delivered to the load is a function of the waveform and the load resistance.

Now, imagine replacing v(t) with a DC voltage (called Veff). In that case, the average power is simply Pavg = [(Veff)^2]/R.

What you want to do is find the value of Veff that will deliver the same average power to a load, R, as the time-varying v(t). You will discover that the value of R drops out of the equation and that Veff is equal to the square root of the mean (the average) of the square of v(t), or, in shorthand, the root of the mean of the square, or RMS; so we generally call the Veff the Vrms voltage, instead.

With this in mind, you should be able to find the RMS voltage of the following voltage signal:

v(t) = V*sin(wt)

See what you get up to this point, and then we can talk further.

3. ### kvnsmith.at.uw Thread Starter New Member

May 16, 2012
4
0
I'd go with V / root2 but it's more of a hunch than a calculation...

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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783
Took a look at the video. The teacher is using phasor notation in the frequency domain. It is standard practice to represent sinusoidal quantities in the time domain as RMS phasors. Since phasors have no "frequency" in the basis frequency domain then one doesn't need to be concerned with sinusoidal terms. Phasors have only magnitude and direction [angle]. This is sufficient to translate any phasor back to the time domain at the basis angular frequency.
Quite often questions are posed without specific reference to a function being a cosine or sine term. For instance a question might simply state that the source is say 120V (RMS) @ 60 Hz. That might be all one is told about the source - not even a source phase angle. Normally the source angle is assumed to be reference of 0 deg unless specifically stated to the contrary.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Interestingly, we use angle in degrees for both time & frequency domain forms. Strictly speaking one should use radians in the time domain form rather than degrees but we seem to overlook the issue and hope people remember to do the conversion when say an instantaneous time value is required.