00:01
This problem uses kirchhoff's current law.
00:04
I've drawn just the outline of the circuit, and i've labeled the nodes a, b, and c.
00:10
I'm going to apply kerchhoff's current law at each one, which says that the sum of the currents in is equal to the sum of the currents out.
00:17
Also going to use oms law, and i know that v0 over here is 4 volts.
00:26
And so i'm going to use that to work backward, backwards to 4.
00:30
Find out what v sub s is over here.
00:34
So i'm going to do kirchhoff's current law.
00:38
If i apply kcl at a, then i see that i have two millimets coming in, and that's equal to i1 plus i2.
00:50
I also know that i1 is going through that one kilo -oom resistor times one kilo -oom equals to v -0.
01:00
So then i do, equals to 4 volts.
01:03
Therefore i1 is equal to 4 millamps.
01:08
So i'll use that to find out what i2 is.
01:12
So i2 therefore is equal to negative 2 milanps.
01:18
I'm going to keep it pointing the same direction and just make sure that i use the negative sign.
01:23
Then if i do kerchhoff's current law at b, then i2 is equal to i3 plus i4.
01:36
So now i'm going to use the voltage drops from b to c, from b to a, from a to c.
01:45
And i'm going to use those expressions, combine them with these, and then i'll be able to work up to the s.
01:55
So a voltage drop from a to c is equal to 2 times i1.
02:02
Plus 1 times i1 so that's equal to 3 times i1 that's just working around the right side of the current of the circuit and so it's equal to 3 times 4 milliamps and so my voltage drop over there is 12 volts i also know that the voltage drop from a to c is equal to voltage drop from a to b.
02:29
I don't know why i wrote an a there.
02:32
So it's equal to the voltage drop from a to b plus the voltage drop from b to c...