00:01
We have a circuit and we would like to determine the current in each branch of the circuit.
00:06
So let's just take a look.
00:08
We have a circuit and there's three branches and i have said that current i1 is flowing through the left segment and current i2 is flowing through the right segment.
00:21
So let's just take a look.
00:23
Let's say that this leftmost branch is branch a and flowing through that will be i1 let's say the middle branch with the 5 ome resistor the 1 omer resistor and the 4 volt power supply is branch b and in branch b we will have current i2 minus i1 and then in branch c which is the rightmost branch with the 3 oom resistor one -on resistor and 12 -volt power supply, we will have current i -2 flowing through it.
01:04
Our goal now is to find current's i -1 and i -2, and in order to do that, we should use kierkov's rules.
01:12
And to do that, you're basically just using omsla again and again and again.
01:17
So let's look at the segment with current i -1.
01:21
So if we follow it up from the bottom left corner we get, i'm going to ignore units now, but this is our first equation.
01:33
We're going to get 8 i1 plus now.
01:41
Now we're in the middle branch.
01:43
We're going to get 5 times i1 minus i2 because, as you can see, i2 is coming up with that middle branch but i1 is going down but we're concerned with i1 right now plus i1 minus i2 again because we have a one on resistor and then we have minus 4 and this all equals zero because any current coming in has to equal the current going out so everything we are doing is going to equal 0.
02:22
Now, for segment 2, let's start at the 4 volt power supply.
02:28
So we are going into the negative terminal without the positive.
02:31
So this will be a positive 4 here.
02:34
And we have here, we have i2 minus i1, because now we're concerned with current 2, plus 5 i2 minus i1 plus 3 i2 plus i2 plus i2 and now it's minus 12 because we're going in the positive terminal out the negative and that is not the way current conventionally flows and again this equals zero so we have two equations and two variables what we're going to do now is solve equation one for either variables i am going to solve it for i2 in terms of i1.
03:27
So let's do that.
03:29
You just want to add up all i1 values and all i2 values.
03:35
And what you should get is 14 i1 minus 4 equals 6 i2.
03:49
Now we can divide by 6 on both sides so that we get i2.
03:57
Is equal to 7 thirds i1 minus two thirds.
04:09
And so this two thirds will be amps, but again, we just don't really care right now because doing it unitless, as long as you're doing it properly, makes your life so much easier.
04:25
All right, now that we have this, we can plug this into this equation...