00:01
Okay, so this is going to take a little bit of work and we're going to have to apply kirchhoff's rules.
00:08
I'm going to start by applying the junction rule at this point right here.
00:13
So we have i18 coming into this junction and we have two currents i12 and i36 coming out of this junction.
00:22
So i'm going to label this equation number one.
00:25
We're going to have i18 is equal to i12 plus i36.
00:34
So i36.
00:36
And so what this is implying is that the current moving into that junction i18 has to equal the currents leaving that junction which are i12 and i36.
00:51
Now let's move into our loop rules.
00:54
I'm going to write two loops, one starting at this 18 volt battery and then i'll write another one that's going to go around this bottom loop down here.
01:05
So i'll write those two equations.
01:06
We could also write a third loop equation that goes around the outer edge, but we're not going to need that one in this case.
01:15
So let's just do the two.
01:19
So to start, i'm going to start at the battery and move with the current.
01:23
So we'll have 18 volts from that battery and then we'll subtract the voltage drop across that 8 ohm resistor.
01:30
So 8 times the current there, which is i18.
01:34
Then we'll subtract the voltage drop across the 11 ohm resistor.
01:40
So 11 times i12.
01:42
We can add that 12 volts from that battery.
01:48
Then we'll subtract the voltage drop across the 7 ohm resistor.
01:53
So that resistance times its current, which is i12 again.
01:58
And then we are coming back up.
02:00
So we're just working our way around and then we hit that 5 ohm resistor.
02:04
So we'll subtract 5 times i18 again and then that brings us back to our starting location of that 18 volt battery.
02:15
So all those voltages added together need to add up to be 0.
02:19
So the voltage drops and the voltage gains all have to cancel each other out over this.
02:25
So let's simplify this equation a little bit.
02:27
We have an 18 and a 12, so those can add together to be 30.
02:31
And i'm actually going to move those to the right side of my equation.
02:37
And i'm just going to flip the signs on everything so that i have my numbers on the right side and my variables on the left.
02:44
And now we'll have two of these i18s.
02:48
So we'll add those together.
02:49
So 8 and 5 give us 13 i18.
02:55
And we have 11 and 7, which gives us 18 i12.
03:02
And that's going to be equal to that 30 volts, the combined 18 and 12.
03:08
So we'll label this equation number two.
03:12
Equation number three we'll get from this loop down here.
03:15
So let's go ahead and work our way around.
03:17
I'm going to start at this junction, although you can start it anywhere.
03:21
Actually, let's start right here at this 36 volt battery just for consistency.
03:26
So starting at the 36 volt battery, we have 36.
03:29
Now we're going to be going against the implied current of i12.
03:33
So we're going to actually add the voltage drop from the 7 ohm resistor, so 7 i12.
03:39
We'll have to then subtract this voltage because we're going against that battery, so minus 12.
03:46
Add this voltage drop again, same reasoning, 11 times i12 in that spot.
03:55
Then we hit this corner, come around, and we'll have a voltage drop across that 5 ohm resistor that's in the direction of the current.
04:02
So that's 5 i36.
04:05
And that again is going to be equal to 0.
04:07
Now we'll do the same trick we did up here with this equation, where i'm going to move the numbers to the right -hand side, the variables on the left -hand side, and we'll just flip the signs on everything.
04:19
So we'll have a positive 5 i36 minus that 18 i12 because we're adding 7 and 11 again.
04:34
And then on the opposite side, we have 36 minus 12, which gives us 24 volts...