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Okay, we're going to use the techniques that you learned earlier in the lecture

today, to look at a much more complicated problem.

And this is actually the last homework problem, and I'm going to leave the bulk

of the work to you. But I want to talk a little bit about it,

and show you how to set it up and how to go about this problem.

This is actually an example of nodal analysis.

Which is a, a, a general technique used to find the currents.

And, and then the voltages in circuits. Now, this particular circuit

configuration is something called a bridge circuit.

it's built out of these two strings of resistors.

And then the midpoints of these two voltage dividers are connected by another

resistor. Now, the values that I put on the

resistors were 1 ohm for these three and 2 ohms over here.

If this one were also 1 ohm, then this would be a very simple circuit to

understand. What happens then is, there's a voltage

up here that's common to both of these branches, and both ends of the branches

are grounded down here. And so if this were, if these were all 1

ohm, then these would just be voltage dividers, and I'd have half of the

voltage up here. Showing up at a, and I also have half of

the voltage up here showing up at b. So then, a and b would be at the same

voltage in that case. And in that case, there wouldn't be any

current flowing here, and the, the, and that simplifies everything quite a bit

because then if I, if there's no current I can just take that out, and then I can

just take the series combination of, of this and this.

And put that in parallel with this, and figure out this resistance, and then the

current is easy to find. But, if I'm a little bit perverse about

this, and I make this resistor have a different value, then that throws

everything out of balance and there's going to be some current flowing here.

In fact, you use this circuit, this circuit is used for measuring

resistances. If I put a resistor in here and I don't

know its value. I can, by measuring this current, I can

figure our this unknown value of this resistor.

And in fact, this is used a lot in sensing circuits.

You might have something who's, a sensor who's resistance changes a little bit.

And you can put that element in this bridge circuit.

And then monitor what's happening to this resistance by looking at this current.

Well, that's an aside, but what I want to show you is how to set this up.

Now, the systematic way to go about this, is I have six independent currents in

this, or six different currents that are all linked together through the set of

equations, but there's six currents in all of the different branches of this

circuit. So, what I need to do is figure out how

to write down six equations that I can then solve.

Now, if I start looking around here and think about how, where can I apply

Kirchhoff's Current Law and where can I apply Kirchhoff's Voltage Law, I can see

where I'm going to get those equations. Now one place is, I can use Kirchhoff's

Voltage Law and go around this loop here. On the left, I can then go around this

loop and around this loop, and that will give me three equations.

Then, I can use Kirchhoff's current law by looking at this node up here.

And, taking the algebraic sum of all the currents going into that node and making

that 0. Then I can do the same thing at this node

and that node. And so, that'll give me six equations,

all together. So, let's take a, take a minute and try

to write down these equations. So let's see I start off on this loop

over here that we'll call that loop a. Let me get myself working here.

So we'll call this A, and we're going to go around this way.

So if I go around that loop a, I'm going to go through the battery first and I

pick up 12 volts, plus 12 volts, then I'm going through this resistor in the

direction that the current is flowing, and so I am going to lose.

Let me erase that and start over again, sorry.

So I'm going to go around this loop I'll call a.

And so I pick up 12 volts, and then I'm going through this resistor in the

direction of the current, so there's a voltage drop.

That is 2 times the current, two i. Now, I'm going through this resistor in

the direction of the current, so there is a voltage drop which is 1 ohm times i2.

And then, to complete the loop, I'm going through this resistor, and there's an

additional voltage drop that is 1 ohm times i3, and those have to add up to 0.

So that's this loop A. Now, if I go over here and go around this

loop, and I'll call that b. So this is A, B, now this gets messy,

because when I'm going around this loop. I'm going through some of these resistors

in the direction of the indicated current.

But when I go through this resistor here, I'm going opposite the direction

indicated. So, voltage, when I go with the current,

it's a negative voltage. If I go opposite the current.

I'm going from the low to the high side of that resistor.

It's a positive voltage. So let's start at this point, A.

And go around and come back to A. So, I'll go up through the, that 1 ohm

resistor that i2 is flowing through. And I'm going to go opposite the

direction, so there's going to be a voltage picked up, which is i2 times that

1 ohm. Now I'm going down through this one ohm

resistor in the direction that i4 has indicated, so that's going to be minus 1

times i4. And then, I'm going through this ten ohm

resistor in the direction indicated by i1, so that's going to give me minus 10

times i1, and that has to be zero. Now to complete our set of equations I'll

go around loop C. So, if I start down here at ground, first

I'm going up through this resistor opposite the direction of i3.

And so, that's going to be i3 times 1 ohm.

Now I'm going through the ten ohm resistor opposite the direction of i1, so

that's going to be plus 10i1, because I'm going counter the direction of the

voltage drop, so I go from the minus to the plus side.

8:12

So this is, this is the plus side, this is the minus side of that resistor.

The current goes from plus to minus so that's that, and then I go down through

this 2 ohm resistor. so I have a voltage drop.

So it's 2 times i5, and that equals zero. Now, those are the 3 KVL, Kirchhoff's

Voltage Law equations. Now, let's go here and look at a.

If I take everything, every current flowing into a is positive, currents

flowing out are negative, and so, at node a I have I2 flowing in.

I have I1 flowing in, and then, I have I3 flowing out.

So those add up to 0, then at B, I have I1 flowing in.

I'm sorry, i4 flowing in, right up here. Then I have i1 and i5 flowing out.

So, this is minus i1 minus i5. Then there's one node left, and that's up

here, this node. We'll call that little c.

And then at node c, I have i flowing in, and then I have i2.

And i4 flowing out. Okay, and so I get those three equations

using Kirchhoff's Current Law. Now, it's all set up.

There's nothing left but some hard work to solve this.

We have the unknowns here are i, and then i1 through i5, and there's six different

variables we're looking for once we found all of these current, it's easy then to

find this voltage. This is the reference, so it's just going

to be this current, i3, times 1 ohm. That's going to tell me this voltage, and

this one is going to be what ever i5 is time 2 ohms, that's 2 times i5 higher

than the ground. And so it's easy to find those.

So the problem that you have to solve then, is to solve this set of equations.

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So if you know some linear algebra, you can do it in that method.

Or you can just do it by substitution. Now, let me give you a hint here.

What you have to do is, to solve a set of equations like this, is you just have to

systematically get rid of variables until you get down to one equation and one

unknown, and you, you solve that one. And then, you work backwards,

subsituiting back in finding the, the, the the variables that you eliminated

along the way. Now the way to start is take a look, look

for the equations, look for the variables that appears in the least number of

equations, and start there. And so I look at and I see i5 only

appears in two equations. So then I can take this equation, I can

solve it for i5. So that's i5 equals i4 minus i1.

Then I can substitute that in up here And work out whatever that is.

So it's i3 plus 10 i1 minus 2 times i5. It's 2 minus 2i4, and then minus 2 times

minus 1, so it's plus 2i1, and that's equal to 0.

So, now I've gotten a new set of equations where I've got this one

equation here that doesn't have i5 in it anymore.

I forget these two equations for now this one and this one.

And so now I've got one, two, three, four, five equations with five unknowns.

And you just keep chipping away at it that way.

eliminating variables til you get down to, to one equation.

So, have fun. This is going to keep you busy probably

until next week. So this is a you know, this is kind of a

sophomore level circuit analysis course that you'd get in a college course.

And it's just finding the solution is a, a testament to your perseverance and

ability to do algebra without messing up. Good luck.