Implications of difference between RMS and mean in rectified and non rectified graphs

Short answer: in most cases RMS values should be considered to calculate power in a component, however if there is a need to calculate power supplied by a DC source, then the mean or DC components should be used.

An important distinction should be made: When I first asked this question I wrongfuly thought that a Multimeter set to AC volts or amps displayed the RMS value of a signal regardless of whether DC was present or not, so when both DC and AC were present, I was confused on which value to use for example to calculate power, instead, when set to AC, a multimeter displays the RMS value of the AC component of the signal only, however, if you want the RMS value of a signal in which both DC and AC are present, then you should measure both the AC and DC component in a multimeter and \$V_{RMS}=\sqrt{V_{DC}^2+V_{RMS_{AC}}^2}\$ should be used. It is obvious that if there is no DC present, the mean value would be zero and the value displayed by the multimeter set to AC is in fact the RMS value of the signal, .

The RMS value of a signal is

\$RMS=\sqrt{\frac{1}{T}\int_{0}^{T} f(t)^2dt}\$

This is the value that should be used, for example in a rectified signal through an LED.

The contribution of both the DC and AC components can be easily seen if the analysis is focused on harmonics, then, power is calculated as:

$$P=V_{DC}I_{DC}+\Re \{\frac{1}{2}\sum_{n=1}^\infty V_nI_n^*\}$$

Where:

\$V_{DC}\$ and \$I_{DC}\$ are the DC voltage and current

and

\$V_n\$ and \$I_n\$ are phasors and include the peak voltage and current of the nth harmonic along with its phase.

In the case where only one frequency is present, then \$P\$ is simply

$$P=V_{DC}I_{DC}+\Re \{\frac{1}{2} V_pI_p^*\}$$

Thus, the power in for example a resistor, is due to both the DC + AC component.

When calculating the power being supplied by a DC source, the DC voltage of the source and current through the source must be considered to calculate the power being delivered by the source, same thing happens with an AC source, but in that case the AC voltage and AC current should be considered.

Regarding current, the RMS value is

$$I_{RMS}=\sqrt{I_{DC}^2+\frac{1}{2}\sum_{n=1}^{\infty}I_n^2}$$

Where

\$I_{DC}\$ is the DC component and \$I_n\$ is the peak value of the nth harmonic, again if only the fundamental is present, the equation reduces to:

$$I_{RMS}=\sqrt{I_{DC}^2+\frac{1}{2}I_p^2}$$

The RMS voltage is calculated in a similar way, thus, in general, in order to calculate power in a component in which both the DC component and the AC component are present, we must consider the RMS value.

Consider the following example of 2 resistors in series, there is also a 10V AC component on top of a 12V DC component feeding the circuit, I also added a power meter and a current-voltage probe.

The Peak voltage is clearly half of the peak to peak voltage, so

$$V_p=9.94/2=4.97V$$

The DC voltage is

$$V_{DC}=6V$$

The RMS voltage is:

$$V_{RMS}=\sqrt{6^2+\frac{1}{2}4.97^2}=6.95V$$

Which agrees with the value displayed in the yellow box in the picture

The current can be calculated the same way, its value is

$$I_{RMS}=6.95 mA$$

The power is simply \$P=V_{RMS}I_{RMS}=48.3mW\$ which agrees with the power meter, (Note: I have noticed that in Multisim the voltage and current values displayed by the probes are not 100% accurate, as opposed to the values displayed by the Multimeter which are more precise, this is why theres a slight difference between the calculated power and the power displayed by the power meter)

Note that the power could have been computed using \$P=V_{DC}I_{DC}+\Re \{\frac{1}{2} V_pI_p^*\}\$, and the results would be the same.

Let's clarify what "effective" voltage means. First of all let's note that this is only really used to describe time-varying periodic voltages. We can imagine taking that voltage and driving across some resistor, R. Since this is a periodic wave it will on average consume some set amount of power.

If we take the same resistor and drive a particular DC voltage, we will consume the same amount of power. This voltage is the Effective Voltage. It is independent of the resistor used. Different resistors will give different power consumption, but for a given periodic voltage input, the effective voltage is the same.

If we go to work the problem out for determining what this effective voltage is for any wave, it effectively comes down to performing the following steps. I recommend working this out from V^2/R to understand why these steps are correct.

1. Divide the period up into a bunch of small samples.
2. Square the voltage value at each sample.
3. Average up all these squared values.
4. Take the square root of this average to get an approximation of the effective value.

As you do this with more and more samples (i.e. smaller time between each sample), you approach the effective value. This sequence is called the root-mean-square, or RMS. In other words, RMS Voltage is Effective Voltage.

Now, here is where people get confused. If you go and do the math, you can calculate analytically that for a sine wave $$V_{RMS} = \frac{V_{peak}}{\sqrt 2}$$ This is only true for a sine wave. For most other waves, you simply have to go do the math.

However, for some waves, like a full- and half-wave rectified sine waves, you can use the definition to determine the RMS voltage more simply.

For a full wave rectified wave driven across a resistor, it can be realized that a resistor doesn't care if a voltage is positive or negative, just what the magnitude is. The power is the same as a regular sine wave, therefore the RMS voltage is the same.

For a half wave rectified wave driven across a resistor, again, it can be worked out that the power consumed is 1/2 of the typical sine wave. The power is 1/2 the regular sine wave, but voltages are squared to get power, so:

$$\frac{P_{sine}}{P_{half}} = 2 = \frac{{V_{{RMS}_{sine}}}^2}{{V_{{RMS}_{half}}}^2}$$ which gives $$\frac{V_{{RMS}_{sine}}}{V_{{RMS}_{half}}} = \sqrt 2 => {V_{{RMS}_{half}}} = \frac{V_{{RMS}_{sine}}}{\sqrt 2} = \frac{V_{peak}}{2}$$

A bit more complicated, but it's easier than calculus.