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I am trying to measure RMS value of full wave rectifier but I don't know why, I see two different RMS value. See the image below enter image description here

According to voltage prob, RMS value is 3.25V and according to multimeter, RMS value is 1.69V Also Vdc = 2.77 but I know Vdc = Vrms. What is the problem ? Why I see different values ?

I Changed place of ground but result were not changed. I didn't connect osolloscope but result was same.

RMS value should be Vmax/sqrt(2) and this value is 4.84/sqrt(2) = 3.422 but which multimeter shows is half of the result.

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  • \$\begingroup\$ Piko if you are done with this question and have a satisfactory answer, site etiquette suggests you should upvote useful answers and formally accept the single answer that is most useful to you. This applies to all your questions previously answered. Take the tour to understand why and to see how this is done. \$\endgroup\$
    – Andy aka
    Commented Dec 12, 2020 at 15:57
  • \$\begingroup\$ @Andyaka Actually I don't have a satisfactory, you and someone said something and I have been researching about what you said. For instance: Why we used V_{RMS} = \sqrt{V_{AC}^2 + V_{DC}^2} I don't know this formula. \$\endgroup\$
    – Piko
    Commented Dec 12, 2020 at 19:08

2 Answers 2

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  • The multimeter shows the RMS \$\color{red}{\boxed{AC}}\$ content at 1.699 volts AC (VAC).
  • The V(dc) value (as reported in the question) is 2.77 volts.

$$V_{RMS} = \sqrt{V_{AC}^2 + V_{DC}^2}$$

$$V_{RMS} = \sqrt{1.699^2 + 2.77^2} = 3.2495\text{ volts}$$

This matches the RMS value reported in the question (3.25 volts).

but I know Vdc = Vrms

No, that's not true - Vrms is equivalent to a DC voltage in terms of the amount of power it can deliver to a resistive load. A sinewave has no DC value yet it has an RMS value and that value is certainly not equal to the average or DC value of the sinewave.

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  • \$\begingroup\$ What is difference AC RMS and Vrms also where does Vdc come ? We call RMS value as effective DC voltage, don't we ? How does simulation calculate the Vdc ? \$\endgroup\$
    – Piko
    Commented Dec 12, 2020 at 11:19
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    \$\begingroup\$ The sim will average the voltage waveform to get the DC (average) value. That isn't however the same as the equivalent DC value of an RMS waveform in terms of heating a resistor. That may be your confusion. \$\endgroup\$
    – Andy aka
    Commented Dec 12, 2020 at 11:24
  • \$\begingroup\$ Where does come Vrms formula ? Also how is RMS AC calculated by Multisim ? \$\endgroup\$
    – Piko
    Commented Dec 15, 2020 at 8:19
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    \$\begingroup\$ @Piko you'll have to ask Multisim how they calculate RMS. I don't work for them and I don't know their secrets. The VRMS formula is craved in stone like here. If you are done now? \$\endgroup\$
    – Andy aka
    Commented Dec 15, 2020 at 9:48
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    \$\begingroup\$ I guess yes, I am done. Thank you so much. \$\endgroup\$
    – Piko
    Commented Dec 15, 2020 at 14:48
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The formula Vrms = Vpk/sqrt(2) applies only to sine waves. The formula does not apply any more if the waveform is not a sine wave, and rectified sine wave is not a sine wave any more.

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  • \$\begingroup\$ so why prob shows a value which is calculated with sin wave RMS formula. Also I researched and there are same formula for full wave rectifiers. \$\endgroup\$
    – Piko
    Commented Dec 12, 2020 at 10:03
  • \$\begingroup\$ The multimeter is on AC. It won't show DC component so it is not RMS. And rectified sine has DC component. \$\endgroup\$
    – Justme
    Commented Dec 12, 2020 at 10:11
  • \$\begingroup\$ but I think it is not excatly DC. if it is DC, shouldn't be sin^2(wt) DC as well ? İf it is not RMS value, what is it ? Also I can give you an expamle: When I try to measure voltage of a battery by AC multimeter I see zero and battery is more DC than rectifier wave. And when I connect and disconnect the prob of AC multemeter to battery in definite time, I see voltage because I give it a frequency İf this wawe is also DC, I should see zero ? \$\endgroup\$
    – Piko
    Commented Dec 12, 2020 at 10:23
  • \$\begingroup\$ It will have some DC component, no matter which way you look at it. You can verify that in many ways, for example the waveform is not symmetric around 0V but always above it so it must, by any observing method, have a DC component. The AC ripples wave around that DC component. A battery has no AC ripples around the DC component. \$\endgroup\$
    – Justme
    Commented Dec 12, 2020 at 10:53
  • \$\begingroup\$ My brain is really mixed. I have never seen a DC component at AC circuits. What do you mean ? What is difference between AC and DC components. Also our teachers have never said something like you said. \$\endgroup\$
    – Piko
    Commented Dec 12, 2020 at 11:16

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