Welcome back to our continuing series in linear circuits.
Again, I'm Nathan Parrish. And today we're talking about our first
real circuit elements, resistors. In this class we're going to introduce
resistors and how to analyze them in circuits.
Putting it in context, we've already talked about a little bit of how within a
device we see the voltage and the current relate to each other.
Then we talk about Kirchhoff's laws, which allows us to see how devices work
together. Now when we have real resistors, we can
put these two ideas together and start doing some basic simple analysis.
So, again, from our previous class, we covered Ohm's law and then we covered
Kirchhoff's laws, which are these single device, as well as multiple device
interacting properties. The objective to this lesson is to apply
those two laws, Ohm's law and Kirchhoff's law to simple resistive circuits, then
calculate an equivalent resistance of resistors when their placed in Parallel
or series and then we are going to do an example, where we can see that, you can
apply these laws, sucessfully to solve a complete solution and find an equivalent
resistance. So to review, first of all, Ohm's law.
They are going from the plus to the minus V equals i times R.
If you flip your references, multiply by negative 1, Kirchhoff's Current Law, you
sum all of the currents coming in to have to equal the currents going out, and then
Kichhoff's Voltage Law, if you make any closed loop and sum up all the voltages
along that closed loop, Then the sum has to be equal to zero.
You end up back where you started. If I stick resistors in series, we can
find that, our equivalent resistance by simply adding the 2 resistors.
And to show why this happens, there's a derivation here that shows that, because
they're in series. They have to have the same current, and
using that, you can then factor out to discover that R is equal to R1 plus R2.
This brings us to the idea of a voltage divider.
What I've placed two resistors in series, in this configuration, and I've put some
voltage across them, we'll see some voltage dropping in the first resistor
and some voltage dropping in the second resistor.
And to see exactly how that interaction works and to see what the voltage is on
that center node in between them, we can use this analysis called the voltage
divider. And essentially what it gives us is if i
want to know the voltage v2 Which is across this resistor.
I find the resistance of this resistor, r2, here.
And then I take the sum of the 2 resistances, r1 plus r2.
So, if r2, were, for example, 3 ohms. And r1 is 1 ohm.
3/4 of the voltage drop would occur across r2.
1/4 of the voltage drop would occur across r1.
And similarly, you can calculate v1. But just using an r1 instead of an r2.
When I place them in a parallel configuration.
Remember that, now, the voltages have to be equal, due to Kirchoff's voltage law.
Doing a little bit of analysis, we discover that the equivalent resistance.
Of resistors placed in a parallel configuration, is the sum of the inverse
of the resistances, and then you take that quantity and invert it.
And when you're doing the analysis for this, it sometimes ends up being easier
to analyze using the conductance rather than the resistance, and then change your
answer when you get to the end. And so that's the way that the derivation
is presented here. And you can take a closer look at that if
you have questions on, on where this calculation came from.
Like we had the voltage divider, there's also a current divider.
This comes from Kirchoff's current law. Because the current coming into this set
of resistors. Has to be split between all of the
different paths or I is equal to i1 plus i2.
When we use that, we can use the, the previous calculation of the currents that
are going i1 and i2 from before. And plug them into to discover that i1 is
going to be equal to r2. Divided by R1 plus R2 plus i.
Be careful, because this is i1 and here we're using R2 in the numerator.
This is i2, now we're using R1 in the numerator.
If you have more than 2 elements, these equations change a little bit.
Normally, we're only going to do two at a time, but if you're interested, you can
do the same derivation and discover what these equations would be if you had three
or more resistors. Finally, we're going to do an example to
apply the things that we've learned. Here we have a 7-ohm resistor and a 5-ohm
resistor. And we see that these are now in a series
configuration right now. The six ohm resistor is then in parallel
with that combination. And then this eight ohm resistor is in
series with the whole thing. So to be able to solve this, what we're
going to do is, we're going to do them each piece at a time.
So first of all, this series is very easy to calculate.
The equivalent resistance is going to be just 7 plus 5 is 12 ohms.
Because, R is going to be equal to R1 plus R2.
Now, that combination is in parallel with the 6 ohm resistor.
So, to do that will do one over 6. Plus 1 over 12.
Then we will invert that whole result. So that's 2 12ths plus 1 12th is 3 12ths,
so 1 4th, and invert, giving us 4 Ohms. because in this case, R is equal to 1
over R1 plus 1 over r2 inverse. That gives us the equal resistance of
that whole set. Finally we have this 8 ohm resistor which
is now in series with this combination, so the final result is going to be the 8
ohm resistor plus our 4 ohm resistor. Giving 12 ohms.
So what's the significance of this? Well if I have all of these resistors, I
could replace the whole network with a single resistor that has a 12 ohm
resistance. So this could be applied both ways.
If you have a 12 Ohm resistor, you could replace these four resistors with a
single resistor, or if you don't have the resistor you want, you can use
combinations of resistors and parallel series to get the desired quantity.
To summarize, we introduced resistors as circuit elements, and then we showed how
they can be combined when they're in a series Or in a parallel configuration and
we found those equivalent resistances by using these laws and we found that we can
use them successively. applying them again and again until we
get the desired result. In the next class there will actually be
a demo showing that all of things that we've been talking about actually works
in real life and we'll be able to see the way that the voltage is dropping it would
cross through this resistance. And that will be demonstrated.
If you have any questions on the stuff that was covered in this in this lesson,
as always go to the forums. And until then, see you later.