00:01
Okay, so you need to find the potential difference between a and b.
00:03
Let's say that r is 5 oms and the voltage in your battery is 8 volts.
00:09
If you have different numbers, you can just plug in the different numbers as we solve the equations and all will be okay.
00:17
So first, we are going to let i1 point downwards.
00:27
I is going to point up and i2 is going to point down.
00:38
So this choice is slightly arbitrary.
00:40
You can choose it how you want, but this is the way i've solved this problem.
00:44
So i, let me use lowercase, i is equal to i1 plus a2.
00:53
That is the node law.
00:55
So both of these currents are coming into that node and that current is leaving.
00:58
So i1 is equal to, sorry, i is equal to i1 plus a2.
01:02
So in this left loop, let's see what this left loop first, we have epsilon plus 2r, i1, plus i 3 r i so the current going through this resistor is i minus 2 epsilon because here if we hit the negative it's positive if we hit the positive it's negative and then plus 2 r i 1 and this is all equal to 0 so we can say that epsilon minus 2 epsilon is negative epsilon so we're going to move that over added to both sides and that's epsilon which is 2 r i1 plus 3r i plus 2 r i1 so that's 4 i1 so epsilon which is 8 so we're going to plug in numbers 8 is equal to 2 times 5 plus 2 times 5 is let's let this be three just for the sake of it will make the number slightly easier to multiply and since yours didn't have numbers that is an okay change to make so two times three is six and then two times three is six so this is twelve i one plus three times three is nine so nine i.
03:12
Let's save this because this is going to be important...