Why are RMS values considered DC equivalent?

Question

RMS is defined as the AC equivalent voltage that produces the same amount of heat or power in a resistor if the same is passed in the form of a DC voltage to the resistor. But shouldn't the power in AC change continuously due to change in voltage and current and hence producing varying power in the resistor as opposed to the DC circuit where a constant power is generated. I am confused so please help me.

Answer 1

The powers are equal if you consider the AVERAGE power (time average). Many of the other answers have kind of taken shortcuts without explaining all the conditions that must apply for the shortcuts to be legitimate. And you yourself have some subtle mistaken assumptions built-in to your question. If you are an EE student, you should read the rest of this answer.

RMS is defined mathematically as the root of the mean of the square of a function. If the function is periodic (repeats itself) then generally, the mean calculation should be over an exact number of cycles. The function could be anything, and it doesn't need to be periodic. This is the definition of RMS. It has nothing to do with DC or voltage or current at all. In fact, it is often used in statistics.

Instantaneous power in a load is simply instantaneous current multiplied by instantaneous voltage: \$P = V \cdot I\$.

Average power is calculated by averaging the instantaneous power. For repetitive waveforms, the average can be performed over exactly one cycle (or any integer number of cycles). For non-repetetive waveforms, the average must be performed over the entire waveform, or "for a long time." Everything I have written so far is true in a fairly general way. It does not depend on any details about how the voltage or current waveforms look. You can calculate the average power of ANY waveform if you average instantaneous power over a cycle. You can calculate the instantaneous power of any waveform if you know voltage and current.

For DC circuits, it so happens that average power is just \$V \cdot I\$.

In the special case of the sinusoidal voltage applied to a resistive load, \$P_{avg} = V_{rms} \cdot I_{rms}\$, where \$P_{avg}\$ is average power. You can prove this, if you want, by doing the rms calculation over one cycle of a sinusoid.

But, if the load is not resistive, then that equation is not true. If the load is resistive but the voltage is not sinusoidal, then the equation is true, but the RMS voltage will not be equal to \$V_{peak} / \sqrt 2\$, as it is with a sinusoid.

There is one more thing worth mentioning. If the voltage is sinusoidal, and the load is reactive (inductive or capcitive), you can still calculate power if you know something called the "power factor."

For this special case, \$P_{avg} = I_{rms} \cdot V_{rms} \cdot P\!F\$ (where \$P\!F\$ is power factor, and \$P_{avg}\$ is average power).

As far as average power goes, it is often the case that average power is more important than instantaneous power. In general, this is true when the thermal time constant is much longer than the electrical period of the AC waveform. If you look at a high-speed video of an incandescent lightbulb powered by AC, you will see that its brightness does vary a bit as the AC waveform changes, but, because the filament takes some time to heat up and cool down, the perceived brightness of the bulb is based strictly on \$V_{rms} \cdot I_{rms}\$. The mass of the lightbulb itself averages the power somewhat. And your eye averages out whatever ripple remains.

If the filament were very, very tiny, it might not have enough mass to average out the power, and its brightness would vary all the way from near zero to full brightness.

I hope this clears up most of your confusion.

Answer 2

Average power is what gives rise to a sustained heating effect: -

Power is the instantaneous multiplication of v and I.

If we take \$I\$ to be \$v/R\$, then power is \$\dfrac{v^2}{R}\$.

And, average power is the mean of \$\dfrac{v^2}{R}\$.

If we then say that R = 1Ω (just for convenience) we can say:

\${\rm Average\ power} = {\rm mean}(v^2)\$

Then it follows that if we take the square root we get RMS voltage

Answer 3

But shouldn't the power in AC change continuously due to change in voltage and current and hence producing varying power in the resistor

Yes, the instantaneous power in a non-constant voltage/current is not constant.

But in your definition an important adjective is missing. Time Average. You must consider the average over time of the electrical power:

• in the period, for periodic waveforms,
• in the entire signal duration, for arbitrary waveforms.

Answer 4

Integrated power is 'easy' to measure as a consequence of the heating effect. One of the most accurate ways of measuring energy is by measuring the resultant temperature rise.

An AC signal does vary continuously, but the instantaneous information is typically hard to make sense of - it doesn't relate to anything. In all the contexts I can think of which are not quantum/semiconductor effects, what is interesting is the 'average over some time period'. (The peak voltage can be important in other contexts, as noted in comments.)

For an AC signal, you would normally want to average for at least one cycle (otherwise you get a different result).

RMS of a voltage translates directly to being equivalent to the DC voltage if you're considering power dissipation across a resistor. Since this is frequently useful, its what we conventionally use to measure AC - but isn't the only factor which will be important in any specific scenario.

Answer 5

The RMS value is obtained as follows:

(1) The square of the waveform function (usually a sine wave) is to be determined.

(2) The function resulting from step (1) is averaged over time. This is the point where your confusion commes from

(3) The square root of the function resulting from step (2) is found.

Answer 6

Simple example: forget about the RMS or any other terms. Lets us do a simple experiment. Take 230 V DC voltage, and apply it to a heater. Measure the temperature it gives, say T1. Now, to the same heater, apply AC voltage. How much voltage? Until temperature reaches T1. Measure the AC voltage using a Multimeter. You should see 230 V value.

Answer 7

The RMS value of a signal v(t) is,

$$v_{rms} = \sqrt{\frac{1}{T}\int^{T/2}_{-T/2} v(t)^2dt}, $$ where T = time period of the signal v(t).

This is the mean squared value of the signal and its square root is defined as root mean squared value of the signal (RMS).

But if this signal is passed through a resistor R, we get the power dissipated in one period is:

$$Power = \frac{1}{T}\int^{T/2}_{-T/2}i(t)v(t)dt = \frac{1}{RT}\int^{T/2}_{-T/2}v(t)v(t)dt.$$

Thus, power dissipated is equal to: $$Power = v_{rms}^2/R$$

Thus, if we have a DC signal of value \$v_{rms}\$, it will dissipate the same power as the signal \$v(t)\$ when passed through any resistor.

Answer 8

Say you have three voltages \$V_1\$, \$V_2\$, and \$V_3\$. The average power shall be $$P_{\text{av}} = 1/3\left[\frac{V_1^2}{R} + \frac{V_2^2}{R} + \frac{V_3^2}{R}\right].$$

If we define \$P_{\text{av}}\$ as some voltage \$V\$, let's call it \$V_{\text{rms}}^2\$ over \$R\$, then

$$\frac{V_{\text{rms}}^2}{R} = 1/3\left[\frac{V_1^2}{R} + \frac{V_2^2}{R} + \frac{V_3^2}{R}\right].$$

Canceling \$R\$ we get $$V_{\text{rms}}^2 = 1/3\left[(V_1^2) + (V_2^2) + (V_3^2)\right],$$

or $$V_{\text{rms}} = \sqrt{1/3[(V_1^2) + (V_2^2) + (V_3^2)]},$$

which is the definition of \$V_{\text{rms}}\$. The interpretation of \$V_{\text{rms}}\$ is the equivalent DC voltage resulting in the same average power.

Answer 9

Why are RMS values considered DC equivalent?

I would say it's just the opposite: RMS is used and intended to be comparable with the DC equivalent. So RMS is considered DC equivalent because it is its exact purpose, by definition.

As AC signals vary in time, they have minimums and maximums (voltage, current, power). One can consider the maximum, and often it has to, but that would not say much in other situations. RMS instead is a chosen value, comparable to DC situations.

For example: if I have a heater of 100W powered by DC, I know that an AC heater of 100W (RMS!) will heat just the same.

Another example.

My electric heater is 230Vac, 800 watts. Its current is 800/230 = 3.47 A. Its resistance is 230/3.47 = 66 ohms.

When powered with 230V ac mains, the voltage varies from 0 to |325| volts (100 times per second), and the current follows the voltage and varies from 0 to 325/66 = 4,92 A. The peak power so is 325 x 4.92 = 1600 watts!

But those 1600 watts are consumed for very little time, the majority of the time the instantaneous power is quite less, even zero. The net result is that the average power, every 1/100 of a second, is 800 watts.

I would obtain the same results if I powered my heater with 230V DC: 230 x 230 / 66 = 800 watts. And this is possible because the RMS values are intended for that, to give the same result as the signal was DC.

Notice that the peak power vs. RMS power is exactly double, and this is because Vrms and Irms are 1 / root(2) of their respective peaks. Multiplying root(2) by root(2) gives 2. These results are valid for a sinusoid, not for every waveform.