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I am relatively new here and I am confused as to the difference between Vrms and Vm. I would be obliged if someone can explain. (This in relation to 3-phase circuits would be even better)

My shot at learning, but not many results on Google that I can understand.

Reference

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    \$\begingroup\$ What is "Vm"? Define the term or show us a context it is used in. \$\endgroup\$ Commented Apr 8, 2013 at 16:52
  • \$\begingroup\$ This is a good starting tutorial for beginners: youtube.com/watch?v=wnUc5EpgQPM \$\endgroup\$
    – hassan789
    Commented Apr 8, 2013 at 16:55
  • \$\begingroup\$ @hassan789 starting tutorial for beginners is much of a muchness. Just kidding. :-) \$\endgroup\$ Commented Apr 8, 2013 at 17:06

2 Answers 2

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RMS

The RMS value of a waveform is the DC-equivalent voltage. It means, that if you have a sine wave with an RMS value of 10 volts RMS, in order to deliver the same power via DC voltage, you would need 10 volts DC.

Don't confuse the average magnitude with the RMS voltage; Vav does not equal Vrms. In fact, technically, the average voltage of an unshifted sine wave is 0 V.

Vm

Vm generally refers to the peak/max voltage on your waveform.

Graph of AC voltage showing maximum, average and RMS values

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    \$\begingroup\$ Simple and understandable explanation. But what is Vm used for? \$\endgroup\$ Commented Apr 8, 2013 at 17:09
  • \$\begingroup\$ What would the phase circuit voltage be in Vrms or Vm? I have seen many people divide V by sqrt(2) and I by sqrt(2) when calculating the power, while others just leave it as is and then solve for the power. \$\endgroup\$ Commented Apr 8, 2013 at 17:11
  • \$\begingroup\$ The Vm are used when you want to know the max voltage of a waveform. In phasor notation, you are using the Vm as the coefficient. When you are calculating power, you use RMS values. (Note: Dividing by sqrt(2) is the same as multiplying by 0.707) \$\endgroup\$
    – hassan789
    Commented Apr 8, 2013 at 17:14
  • \$\begingroup\$ Thank you again, but the internet seems to be contradicting itself when you look at this document ece.msstate.edu/~donohoe/ece3414three_phase_power.pdf. On page 2, you can see that they described the voltage as in the form of Vrms already. So, does that mean phaser notation is in Vrms instead of Vm? \$\endgroup\$ Commented Apr 8, 2013 at 17:18
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    \$\begingroup\$ This is a bad answer, especially as it relates to Vm, since that was never defined. You are assuming it is the maximum positive voltage, but it could be the mean or something else. Answering without getting the definition first, or at least stating the ambiguity and what you assumed, is irresponsible. This could give the OP completely the wrong idea of Vm. \$\endgroup\$ Commented Apr 8, 2013 at 20:47
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Let's start with this: -

If you had a DC voltage of 10 V and a 1 ohm resistor, the power dissipated in the resistor is 100 W because: -

$$P = \frac {V^2} R$$

The RMS value of 10 VDC is 10 VDC — it's the number you use to calculate power in DC circuits.

In simple sinusoidal AC circuits, if you have a peak value of 10 V, you will find that it dissipates in a 1 ohm resistor somewhat less that 100 W.

It will dissipate 50 W — and if you reverse the process to work out what peak sine voltage would cause it to dissipate 100 W you'll find it to be 14.14 V (approx).

Most engineers are so well-versed in doing this that it may seem, to the uninitiated, that there is no-theory behind it. I acquired this from the web: -

enter image description here

What you refer to as "Vm" I assume is an abbreviation for "Vmax" which I usually refer to as "Vpk".

Regarding 3-phase circuits, there is no fundamemtal difference except you are calculating power three times; one for each phase. If you have 3 resistors in a delta connection then, the power is the RMS Line voltage squared and divided by the delta resistor to which that line voltage is across. Repeat for the two other line voltages and sum all three.

That gives you 3-phase power.

If you have resistors in star-formation, you must use the phase RMS voltage (line divided by \$\sqrt 3\$) and calculate the 3 individual powers. Add them together to get power.

If the loads are imbalanced and there is no neutral connection, then there is more math, but I hope you get the picture.

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  • \$\begingroup\$ You can't know that Vm is Vpk since he hasn't defined it and it's not a standard term. At best, you can guess that's what it is. \$\endgroup\$ Commented Apr 8, 2013 at 17:11
  • \$\begingroup\$ @OlinLathrop OK I'll modify my answer \$\endgroup\$
    – Andy aka
    Commented Apr 8, 2013 at 17:16
  • \$\begingroup\$ I am also referring to your explanation as well. It is quite good now after your edits. \$\endgroup\$ Commented Apr 8, 2013 at 17:31
  • \$\begingroup\$ @OlinLathrop: Indeed, my first guess was that Vm was arithmetic mean (average) voltage; it may well be "max", but it's not clear whether that should refer to "largest magnitude" or "most positive". \$\endgroup\$
    – supercat
    Commented Apr 8, 2013 at 19:48

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