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The mean value of alternating current comes to be zero because of half of the cycle being positive while the other half negative. so, take the root means square value of Alternating current given by:

$I_{rms} = 0.707\times I_{max}$

But why do we take "rms" for direct current? if we take the simple average of direct current, we would come up with a value which is not zero (it is zero in AC).

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See also… because the answer is there too. – Rob Jeffries Nov 15 '14 at 22:39

The notion of RMS voltage originated from electrical engineers trying to calculate the power dissipated from a resistive element. copying from wiki

Let's assume that the average power dissipated through a resistor $R$ be $P_{avg}$. Then,

$$P_{avg} = \langle \frac{v(t)^2}{R} \rangle$$, where $\langle f \rangle$ is the average value of function f.

$$P_{avg}=\frac{1}{R} \times \langle v(t)^2\rangle$$

But, we also have $V_{rms} = \langle v(t)^2\rangle$ by definition of RMS value. Hence,

$$P_{avg}=\frac{1}{R} \times V_{rms}$$

RMS voltage is just a representative value of voltage which gives you the average power in a resistive load.

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i'm trying very hard to understand your question but i still dont get it that why do we take RMS value for DC current? – Rafique Dec 4 '12 at 15:08
the point which I'm making is that electical engg. need some representative value for AC current/voltage for resistive load. It's just coincidence that this representative value is same as root mean square value of voltage/current. – Vineet Menon Dec 5 '12 at 9:00

Keep in mind that the root mean square is slightly different than simply the arithmetic mean value (and in fact will always be equal to or greater than the mean value).

The reason that the RMS value is calculated for direct current so often is that it is used in various calculations such as average power dissipated given a time-varying current or voltage, and we find that it works generally since it correctly solves for things such as power dissipated due to alternating current as well, where a standard arithmetic mean value would not give an accurate answer.

For a bit more clarification, see what Wikipedia has to say on the matter.

Hope this helps!

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AC performs work when the voltage is negative, as well as when it is positive. This means that the "average" is an invalid measure for current or power. Instead, RMS is used.

With RMS the quantity is first squared, to "flip" the negative to positive. It is then integrated, and finally the square root is taken to remove the "error" caused by squaring.

If you perform this operation with a DC voltage, the result is simply the DC voltage. If you do it with a sinusoidal voltage, you end up with 0.707 times the peak value (amplitude) of the sinewave. With other waveforms the result will differ. For some examples, see the Wikipedia entry.

EDIT To clarify the mysterious conversion factor for sinewaves, $\frac{1}{\sqrt2} = 0.707$. This is also the factor you'll notice in the Wikipedia link.

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as per your argument, wouldn't it be better to take a modulo average?? $|v|$ – Vineet Menon Dec 4 '12 at 5:20
It is actually $\sqrt{2}/2=0.707$ – Jaime Dec 4 '12 at 6:28
The problem with that is that AC is not steady like DC. RMS allows for the constantly changing voltage. You need to integrate the waveform rather than average it. @Jaime You're right of course, but then, so am I. Remember that √2/2 = 1/√2 – hdhondt Dec 5 '12 at 22:40
Must have read your answer too fast, I recall having read $\sqrt{2}=0.707$... – Jaime Dec 5 '12 at 22:51
You can only talk about RMS values for DC if the DC is intermittent – Dirk Bruere Nov 15 '14 at 22:01

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