What makes it a good idea to use RMS rather than peak values of current and voltage when we talk about or compute with AC signals.

  • 4
    $\begingroup$ I don't know why the question was downvoted, it seems a perfectly reasonable question to me. $\endgroup$ – John Rennie Oct 26 '12 at 16:04
  • 1
    $\begingroup$ @JohnRennie Check the edit history: the original text was confusing, but think that this was the question that anikumar had in mind. $\endgroup$ – dmckee --- ex-moderator kitten Oct 26 '12 at 18:37
  • 4
    $\begingroup$ Thank you for editing my question , Actually this is the question in my mind but I've some communication problem how to ask it $\endgroup$ – anilkumar Oct 29 '12 at 4:48
  • 2
    $\begingroup$ @JohnRennie because it's a "grubby" engineering question, as Gell-Mann would say. $\endgroup$ – John McVirgooo Nov 1 '12 at 22:31

Attempts to find an average value of AC would directly provide you the answer zero... Hence, RMS values are used. They help to find the effective value of AC (voltage or current).

This RMS is a mathematical quantity (used in many math fields) used to compare both alternating and direct currents (or voltage). In other words (as an example), the RMS value of AC (current) is the direct current which when passed through a resistor for a given period of time would produce the same heat as that produced by alternating current when passed through the same resistor for the same time.

Practically, we use the RMS value for all kinds of AC appliances. The same is applicable to alternating voltage also. We're taking the RMS because AC is a variable quantity (consecutive positives and negatives). Hence, we require a mean value of their squares thereby taking the square root of sum of their squares...

Peak value is $I_0^2$ is the square of sum of different values. Hence, taking an average value (mean) $I_0^2/2$ and then determining the square root $I_0/\sqrt{2}$ would give the RMS.

It's example time: (I think you didn't ask for the derivation of RMS)


Consider that both the bulbs are giving out equal-level of brightness. So, They're losing the same amount of heat (regardless the fact of AC or DC). In order to relate both, we have nothing to use better than the RMS value. The direct voltage for the bulb is 115 V while the alternating voltage is 170 V. Both give the same power output. Hence, $V_{rms}=V_{dc}=\frac{V_{ac}}{\sqrt{2}}=115 V$ (But Guys, Actual RMS is 120 V). As I can't find a good image, I used the same approximating 120 to 115 V.


To further clarify your doubt regarding the peak value, It's simply similar to finding the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ in Cartesian system represented as, (sum of squares & then "root") $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

| cite | improve this answer | |
  • $\begingroup$ I've another question for you , in which cases average value of A.C is useful $\endgroup$ – anilkumar Nov 1 '12 at 11:17
  • $\begingroup$ @anilkumar: Helllo Anil, RMS value is not the average value. For example, 5 volts of DC would have the average value 5. But, 5 volts of AC is not 5. It varies from -5 to +5. The average value is zero everytime. To measure the original effect of AC, RMS value is the most necessary. There are many uses for RMS value of AC like your domestic supply of 220 V. It's indeed RMS value. $\endgroup$ – Waffle's Crazy Peanut Nov 1 '12 at 13:49
  • $\begingroup$ yes I know that RMS value is not the average value, But the average value of Ac sine wave is V(peak) x 0.637 not zero $\endgroup$ – anilkumar Nov 2 '12 at 4:26
  • $\begingroup$ @anilkumar: Ok, Anil. I understand it now. You're talking about the average value of half cycle (not complete). Considering a complete cycle, it'd be definitely zero. In case of half cycle, the expression is the same as you said. But, I'd say that RMS value is mostly preferred than average value of half cycle because RMS would give the equivalent of dc. Have a look here for more info :-) $\endgroup$ – Waffle's Crazy Peanut Nov 2 '12 at 13:42
  • $\begingroup$ Another simple way to look at it, geometrically, is that by squaring V, in effect you flip the negative voltage to positive ((-V)^2 = V^2), then find the average magnitude of these square values. By taking the square root, you get back to an unsquared value that averages the magnitudes, regardless of sign. $\endgroup$ – iSeeker Sep 3 '19 at 17:44

What makes it a good idea to use RMS rather than peak values

The rms value, not the peak value, is the equivalent DC value that gives the same average power.

Recall that power is the product of voltage and current:

$p(t) = v(t) \cdot i(t)$

For a resistor, we have:

$p(t) = R[i(t)]^2$

To find the average power, we must take the time average of both sides:

$p_{avg} = R \dfrac{\int_{T_1}^{T_2} {[i(t)]}^2\, dt}{T_2-T_1} $

You'll recognize the fraction on the right hand side as the mean of the square of $i(t)$.

Denoting $i_{rms}$ (the root of the mean of the square) as:

$i_{rms} = \sqrt{\dfrac{\int_{T_1}^{T_2} {[i(t)]}^2\, dt}{T_2-T_1}}$

we have:

$p_{avg} = R[i_{rms}]^2$

For DC, we have:

$p = R I^2$

So, we see that the rms value of the time varying current produces the same average power, for a given resistor, as a constant current of that value.

This is what makes the rms value "a good idea".

| cite | improve this answer | |

In many applications we are interested in the power. For example your electricity bill is based on the power you consume. For a DC source the power is:

$$ W = VI = \frac{V^2}{R} $$

and for an AC source (assuming a resistive load so the voltage and current stay in phase):

$$ W = V_{rms}I_{rms} = \frac{V_{rms}^2}{R} $$

So using the RMS values makes the power easy to calculate. The RMS values are, in a sense, the equivalent to the values in a DC circuit.

| cite | improve this answer | |
  • $\begingroup$ In case of resistive load ,is the formula W=VI applicable at any instant of voltage and current value? $\endgroup$ – user184271 Mar 2 '18 at 2:04
  • $\begingroup$ @harambe yes. The power I'm calculating is the average power over a cycle, but the instantaneous power is just $W=IV$ where $I$ and $V$ are the instantaneous values of current and voltage. $\endgroup$ – John Rennie Mar 2 '18 at 6:00

Not the answer you're looking for? Browse other questions tagged or ask your own question.