Welcome back. again linear circuits today.
It's going to be maximum power transform, transfer.
This is actually something that you might be able to apply, and use in some of the
things that you actually might be familiar with in electric, electronics.
To do this we're going to be using the feminine equivalent that was presented
before by Dr. Parrish to show the load that gives the
maximum power transfer, we'll also talk about why this is important.
From the previous lesson, there were systematic ways of getting circuit
equations using node-voltage and mesh-current analysis and then Dr.
Parrish presented Thevenin and Norton equivalent circuits where we can take a
whole bunch of components, dependent sources, independent sources, resistors.
Put them all together and get a singe independent source and a single resistor
or a single current source and a single resistor to represent the whole thing.
So today we're going to be doing the next power transfer the next time Wye-delta
and Wheatstone bridge and then finally some demos and applications to conclude
the module. So in the lesson we're going to find the
load resistance that gives maximum power transfer and then calculate what the
power is. How much power is consumed by the load as
well as how much power is being consumed by the system.
So here we have a two terminal linear circuit.
And we know from what Doctor Parrish presented, will be containing the whole
thing and replace it by a thevenin equivalent.
Because we can do this, we're going to just do an analysis using a thevenin
equivalent circuit. And then we'll know that the same results
are going to then apply more generally to any maneur circuit that we might be
working with. But before we get into that we need to
understand how to power for resistance just to make our calculations a little
bit more simpler. So, from what we know, we already know
Ohm's law that v equals i times r. We also know that power is equal to the
current times the voltage. By putting these two ideas together we
discover that power can be calculated by i squared times r.
Or, by taking v squared and dividing it by r.
So, this is what we are talking about when we talk about load resistance.
We have a system, represented by the themanin voltage source in the themanin
resistance here. And then, we connect a load here.
Now, suppose that this load resistance is very, very, very large.
Well, what do we know about that? Well, we know that this resistance is
very very large. First of all, from a voltage divider
stand point, most of this v thevenin is going to be across rl.
But, a very large resistance here is going to give a very, very small current.
And so p is equal to v times i, the result is going to be a small power
consumed by rl. If we look at our thevenin in the same
configuration, our thevenin. The voltage across it is going to become
very small. And this current is also going to become
very, very small. So this resistance is also going to be
consuming very little power. Now suppose that this resistance is very,
very, very small. That this load voltage is going to become
very, very small. Well this voltage here across the
thevenin resistance is going to become fairly large.
But, the current, the power that we are really interested in is this.
So this current here is going to become larger since this is a small resistance.
But the power consumed by the load is going to be fairly small.
So most of the power is going to be consumed by this resistor.
Which isn't really doing what we want either.
So it turns out that there's this sweet spot.
We don't want this resistance to be too large, making the current small and the
power small. And we don't want it to be too small,
making this voltage small. And so the, most of the power is being
consumed here, in this resistor, where the power is being consumed in the
system. So let's derive which resistance we want.
We know that p equals i squared r. And so we can then do this voltage
divider. Finding the voltage drop across this
resistance, to get this equation, uth squared over rth plus rl quantity
squared, and multiplied by rl. And when you do that it turns out that
you can take a derivative and find where this is equal to 0 and do some analysis.
But the, it turns out that the best value that you can pick that's going to
maximize this p, which is the power that's consumed in the load, is equal to
vth squared over 4 times rth. And this happens when rl equals rth.
So the answers to the, the questions that were posed before, is, what load
resistance is the optimal one to use if you want maximum power transfer?
Well, you want it to match the thevenin resistance, the resistance here.
And how much power is consumed by the load resistance in that configuration?
Its calculated by this equation, vth squared over 4 times rth.
It turns out that in that system because now this voltage is equal to this voltage
and obviously being in series their currents are the same.
The same amount of power is being consumed by the load resistance and by
the thevenin resistance. So where can we use this?
If this load resistance is a speaker, for example, and this system here is your
amplifier. You want to choose a speaker that's going
to give maximum power transfer to the speaker, because that equivalate, equates
to more air being moved. More volume.
More power. In your speaker.
So you'd want this, to match this. So you might be familiar with speakers
enough to know that, generally, speakers are labelled as eight ohms.
And you want to make sure that your speakers match up with your amplifier so
that your not wasting extra energy in your system.
That all of the energy well at least as much as you can get is going to the flow
of your system. Now if you pose this question slightly
differently. If we want to know what r thevenin would
get the the most power transferred, turns out that you want to minimize this.
I'm not going to go through the derivation, but it's an excellent
exercise to see why is it that we want to make r thevenin in the smallest possible
to save power and make it, as much of it go to r lotus as we can?
So to summarize, we specified power equations for resistors and we matched
the load resistance for the system resistance to get maximum power transfer.
So as long as you can get a Norton or thevenin equivalent, you can very easily
find out what load resistance you want to use.
We also specified the equation for calculating the maximum power transfer.
In the next lesson we'll talk about a transform that allows us to to work with
resistors that are not in parallel or in series.
They're kind of in the middle ground. As well as analyze what is called the
Wheatstone bridge. Which is a device that is used in
practice for measuring a number of resistances.
Until next time.