Measuring Alternating current/voltage

[alchemist]

When a voltmeter measures the AC voltage in a circuit, does it measures the root-mean-square value of the voltage? Is the same true for ammeters and AC current too?

 

[mgb_phys]

It depends on how expensive the meter is:
Cheap ones measure the peak voltage assume it is a sign wave and multiply by sqrt(2) to give the RMS.
Expensive ones sample the waveform with time and calculate a true RMS reading - these will usually say "true rms" on them.

 

[stewartcs]

Cheap ones measure the peak voltage assume it is a sign wave and multiply by sqrt(2) to give the RMS.

I think you mean multiply by 1/sqrt(2).

Vrms = Vp/sqrt(2)

 

[alchemist]

I see, so you are saying that the value that i read off from the cheaper voltmeter would be the value of the RMS already? or do i still need to multiply the reading i get by 1/sqrt2?

Thanks!

 

[stewartcs]

No, the cheap ones generally measure peak voltage. So, take that reading and multiply by 1/sqrt(2) to obtain the RMS voltage.

 

[f95toli]

No, ALL multimeters display the RMS value. It is just that the cheaper ones are inaccurate when the waveform is not sinusoidal since they just calculate the RMS value from the peak value.
When we talk about AC voltages (or currents) we are usually referring to the RMS value; i.e the mains voltage is 115V or 230V RMS (depending on where you live).
The reason is that we can then get the power from the familiar formula P=V*I (if you instead of the RMS values use the amplitudes you have P=i*v/2)

 

[stewartcs]

Cheap multimeters do not measure true RMS. They measure the average positive voltage of a waveform and scale this value using the square root of two to produce a display value. They may call this value RMS, but it is not a true RMS.

Most of my meters are either true RMS meters or have the ability to measue peak voltage, which of course you can convert to RMS as I indicated before.

 

[f95toli]

Just out of curiosity: Where did you find a meter that displays the peak value?
I most have used seveal dozen different models of multimeters by now (I thknk I have about ten models in my lab) and I have never come across a multimeter that shows the peak value; not even my bench multimeters have that as an option.

Cheap multimeters just assume that it is a sine-wave and divides the peak value by sqrt(2); the only time you need to actually convert anything is if you are using a cheap multimeter and KNOW what kind of waveform you are measuring; then you can sometimes get the true RMS (or the amplitude) by multiplying by a numerical factor.

Anyway, the point is that ALL multimeters display the RMS value; but cheap (i.e. non "true RMS") will simply show the wrong value if you try to measure anything but a sine-wave.

 

[mgb_phys]

Thanks f95toli, I should have made that clear - the cheap meter is displaying RMS but is not necessarily getting it right!

 

[TVP45]

Just out of curiosity: Where did you find a meter that displays the peak value?
I most have used seveal dozen different models of multimeters by now (I thknk I have about ten models in my lab) and I have never come across a multimeter that shows the peak value; not even my bench multimeters have that as an option.

Cheap multimeters just assume that it is a sine-wave and divides the peak value by sqrt(2); the only time you need to actually convert anything is if you are using a cheap multimeter and KNOW what kind of waveform you are measuring; then you can sometimes get the true RMS (or the amplitude) by multiplying by a numerical factor.

Anyway, the point is that ALL multimeters display the RMS value; but cheap (i.e. non "true RMS") will simply show the wrong value if you try to measure anything but a sine-wave.

Keithley makes peak reading DMMs.

 

[stewartcs]

Just out of curiosity: Where did you find a meter that displays the peak value?

Crompton makes a non-RMS voltage meter...model: Analogue 070.

 

[rbj]

Cheap multimeters just assume that it is a sine-wave and divides the peak value by sqrt(2);

Cheap ones measure the peak voltage assume it is a sign wave and multiply by sqrt(2) to give the RMS.

No, the cheap ones generally measure peak voltage. So, take that reading and multiply by 1/sqrt(2) to obtain the RMS voltage.

Cheap multimeters do not measure true RMS. They measure the average positive voltage of a waveform and scale this value using the square root of two to produce a display value. They may call this value RMS, but it is not a true RMS.

actually the cheap voltmeters measure the DC component of the full-wave rectified waveform (the average of the absolute value), assume it's a sine, and multiply by $\frac{\pi}{2 \sqrt{2}}$ to get a pseudo-RMS reading.

 

[mgb_phys]

actually the cheap voltmeters measure the DC component of the full-wave rectified waveform (the average of the absolute value), assume it's a sine, and multiply by $\frac{\pi}{2 \sqrt{2}}$ to get a pseudo-RMS reading.

"After winning the Galactic Institute's prize for extreme cleverness, he was later lynched by a mob of respectable physicists, who finally worked out that what they really could not stand was a smartass."

The inventor of the infinite improbability drive - Hitch Hikers Guide to Galaxy

 

[rbj]

"After winning the Galactic Institute's prize for extreme cleverness, he was later lynched by a mob of respectable physicists, who finally worked out that what they really could not stand was a smartass."

i'm not trying to be a smartass. :-/

 

[mourici]

to get true rms the multimeter must make integration and need about 40 or more data mber per cycle (20msec) . such a meter need a compute (microprocessor) digital scope have a computer in simple multimeters no computer any ac wave is rectified by a diode and a capacitor so the capacitor is charged to max voltage. asuming the voltage is sin on aC scale the meter will show u RMS but valu is correct only for sine wave. u does not need a cakculator meter give valu in rms