>> 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.