>> In a circuit that's being driven by a sinusoidally varying voltage,
the current that's flowing in that circuit is also going to be sinusoidal.
The power delivered to this resistive load which is equal to
the voltage times the current is also going to be varying.
That means, at some points when the voltage and current are at
maximum values you're going to be delivering a lot of power to the load,
and at times when the voltage or the current is zero,
at that instant, you won't be delivering any power to the load.
So then becomes interesting to say,
well is there a way of determining what
the average power is being delivered over one cycle?
Is there are some effective quantity that we can use to
calculate the effective power or
the average power that's being delivered to the load?
We might say, well, let's just take the average of it.
Let's start with a voltage source or some sinusoidally varying thing.
Calculate its average value.
The average value of say
some sinusoidally varying voltage V of t. If V of t is
equal to V_m cosine Omega t,
then, the average voltage is defined as one
over t times the integral from 0 to t,
V_m cosine Omega t, dt.
Without going to bother doing this integration,
we know that when you integrate you are actually
adding up the area under the curve.
In this case going from zero to t,
we're going over one period and adding up the area.
It turns out that in a sine wave,
there's just as much area above the axis positive area as there is area
under the line and thus the average voltage is zero.
Well, that doesn't help us a whole lot because we know
there's power being delivered to
that resistive load but calculating
the average voltage of the average current isn't going to do anything for us.
Instead, we calculate what is known as the root means
squared or the effective voltage.
To do that we start by taking our voltage or current.
We're going to calculate it for both the voltage and the current.
You square that quantity.
Here's the voltage squared.
You'll notice that what was negative is
now been flipped over so it is now positive.
The shape is somewhat different but it's still
ranges from in this case zero to one.
We then average this squared value
and then to bring it back to
the appropriate amplitude or the appropriate units,
we take the square root of it.
In other words, the RMS value or
this effective value of the voltage is found by squaring the voltage,
calculating the average of the square,
and then taking the square root of that.
Let's go ahead and show you what we mean here.
If V of t is equal to V_m cosine Omega,
then V squared of t is equal to V_m
squared cosine squared of Omega t. We're going to take
that and we're going to calculate the average of
that which is done by integrating from
0 to t. Let's bring the V_m squared out in front since it's a constant.
Cosine squared of Omega t dt.
We have to avail ourselves of
a trigonometric identity that says that the cosine squared of
Theta is equal to 1.5 times 1 plus the cosine of 2 Theta.
So to perform this integral,
the trick is to instead of trying to integrate the cosine squared,
we use this identity and substitute in then for this we have,
V_m squared over T times the integral from 0 to
t. We'll bring the 1.5 out in front also,
times 1 plus the cosine of
2 Omega t dt.
Now, when we integrate this over the period,
this term we're actually integrating over two periods.
In the same way that when you integrate it over one period,
the positive area cancel the negative area,
this term here likewise gives us a zero contribution to this integral.
When you integrate one with respect to t,
we adjust t and so we have V_m squared over 2T times
t evaluated from 0 to
T. That then is just V_m squared over 2.
So we squared it,
we calculated the average.
Now all is left to do is to take the square root of it and
we've have in V_rms is
equal to the square root of V_m squared over 2.
If we take that square root of two out from under the radical,
it comes out as 1 over
the square root of 2 times the square root of V_m squared.
Well, that's just V_m.
Thus, we see that the RMS value,
this effective voltage is equal to 1 over the square root of
2 times V_m where V_m was the maximum value, our peak value.
Well, 1 over the square root of 2 that's approximately equal
to 0.707 times V_m.
So the effective voltage is something on the order of pretty close to being
about 70 percent of the peak value 0.707 V peak.
Now then, what does that do for us?
With this effective or this RMS value,
we can now calculate an average power delivered to a resistive load is equal
to V_rms times Pi_rms.
The same expression that we've been used to using all along.
In other words, this RMS voltage and RMS current are the effectively the
same as what you would have if you had a DC source driving it.
To say that even yet a different way,
a DC source with a value of V_rms will deliver the same amount of power to
a resistive load as an AC signal with V_m voltage peak.