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>> In a circuit that's being driven[br]by a sinusoidally varying voltage,

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the current that's flowing in that circuit[br]is also going to be sinusoidal.

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The power delivered to[br]this resistive load which is equal to

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the voltage times the current[br]is also going to be varying.

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That means, at some points when[br]the voltage and current are at

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maximum values you're going to be[br]delivering a lot of power to the load,

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and at times when the voltage[br]or the current is zero,

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at that instant, you won't be[br]delivering any power to the load.

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So then becomes interesting to say,

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well is there a way of determining what

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the average power is being[br]delivered over one cycle?

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Is there are some effective quantity[br]that we can use to

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calculate the effective power or

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the average power that's[br]being delivered to the load?

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We might say, well, let's[br]just take the average of it.

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Let's start with a voltage source[br]or some sinusoidally varying thing.

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Calculate its average value.

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The average value of say

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some sinusoidally varying voltage[br]V of t. If V of t is

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equal to V_m cosine Omega t,

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then, the average voltage is defined as one

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over t times the integral from 0 to t,

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V_m cosine Omega t, dt.

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Without going to bother[br]doing this integration,

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we know that when you[br]integrate you are actually

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adding up the area under the curve.

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In this case going from zero to t,

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we're going over one period[br]and adding up the area.

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It turns out that in a sine wave,

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there's just as much area above[br]the axis positive area as there is area

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under the line and thus[br]the average voltage is zero.

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Well, that doesn't help us[br]a whole lot because we know

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there's power being delivered to

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that resistive load but calculating

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the average voltage of the average current[br]isn't going to do anything for us.

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Instead, we calculate what[br]is known as the root means

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squared or the effective voltage.

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To do that we start by taking[br]our voltage or current.

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We're going to calculate it for[br]both the voltage and the current.

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You square that quantity.

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Here's the voltage squared.

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You'll notice that what was negative is

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now been flipped over[br]so it is now positive.

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The shape is somewhat[br]different but it's still

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ranges from in this case zero to one.

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We then average this squared value

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and then to bring it back to

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the appropriate amplitude[br]or the appropriate units,

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we take the square root of it.

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In other words, the RMS value or

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this effective value of the voltage[br]is found by squaring the voltage,

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calculating the average of the square,

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and then taking the square root of that.

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Let's go ahead and show[br]you what we mean here.

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If V of t is equal to V_m cosine Omega,

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then V squared of t is equal to V_m

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squared cosine squared of[br]Omega t. We're going to take

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that and we're going to[br]calculate the average of

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that which is done by integrating from

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0 to t. Let's bring the V_m squared[br]out in front since it's a constant.

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Cosine squared of Omega t dt.

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We have to avail ourselves of

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a trigonometric identity that[br]says that the cosine squared of

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Theta is equal to 1.5 times 1[br]plus the cosine of 2 Theta.

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So to perform this integral,

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the trick is to instead of trying[br]to integrate the cosine squared,

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we use this identity and substitute[br]in then for this we have,

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V_m squared over T times[br]the integral from 0 to

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t. We'll bring the 1.5 out in front also,

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times 1 plus the cosine of

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2 Omega t dt.

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Now, when we integrate[br]this over the period,

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this term we're actually[br]integrating over two periods.

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In the same way that when you[br]integrate it over one period,

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the positive area cancel the negative area,

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this term here likewise gives us[br]a zero contribution to this integral.

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When you integrate one with respect to t,

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we adjust t and so we have[br]V_m squared over 2T times

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t evaluated from 0 to

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T. That then is just V_m squared over 2.

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So we squared it,

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we calculated the average.

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Now all is left to do is to[br]take the square root of it and

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we've have in V_rms is

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equal to the square root[br]of V_m squared over 2.

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If we take that square root of[br]two out from under the radical,

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it comes out as 1 over

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the square root of 2 times the[br]square root of V_m squared.

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Well, that's just V_m.

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Thus, we see that the RMS value,

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this effective voltage is equal[br]to 1 over the square root of

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2 times V_m where V_m was[br]the maximum value, our peak value.

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Well, 1 over the square root of[br]2 that's approximately equal

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to 0.707 times V_m.

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So the effective voltage is something[br]on the order of pretty close to being

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about 70 percent of[br]the peak value 0.707 V peak.

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Now then, what does that do for us?

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With this effective or this RMS value,

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we can now calculate an average power[br]delivered to a resistive load is equal

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to V_rms times Pi_rms.

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The same expression that we've[br]been used to using all along.

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In other words, this RMS voltage and[br]RMS current are the effectively the

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same as what you would have if[br]you had a DC source driving it.

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To say that even yet a different way,

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a DC source with a value of V_rms will[br]deliver the same amount of power to

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a resistive load as an AC signal[br]with V_m voltage peak.