0:00
All right.
00:01
So in this problem, we have two batteries or two emf sources.
00:06
And we are given all of the resistances, including the two internal resistances of those batteries, as well as the value of the voltage in those.
00:17
And we need to use kirchhoff's rules to figure out the three currents.
00:22
Now, we're given the directions to start with, with i -1 moving to the left in the center, i -2 moving.
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Clockwise around the top and i3 moving counterclockwise around the bottom.
00:36
So the first thing we'll do, or the first thing that i will do, is apply the junction rule and looking at that junction to the left, we have i -1 going into the junction, and that has to equal all the currents coming out of the junction.
00:50
So i -1 has to equal i -2 plus i -3 at that junction point.
00:55
Next, i'll look at the loop on top and do a kerch -off loop, starting with the em.
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Source.
01:02
We have 18 volts on that emf source.
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So that's going to be positive.
01:07
And then minus the voltage drop across r1, a little r1.
01:12
We get 0 .5 times i2, because i2 is the current through here and r1 is 0 .5.
01:19
Then we come around to r1, like capital r1, which is 6.
01:25
So we have minus 6 i1 and then we're coming up around the left side and so we have minus 2 .5 i2 again now at this point let's look at the other loop and then we'll evaluate these equations so the loop on the bottom again starting with the emf source we have 45 volts minus 0 .5 times i3 for the little r and then minus 1 .5 times i3 for r3 and then minus 6 i1 for r1 as we come around counterclockwise.
02:02
And that says equal to zero as well.
02:06
Now let's take a look back at that middle equation there.
02:08
We're going to substitute for i1 with the i2 and i3 from the junction rule equation.
02:15
So we get this equation.
02:17
And i also combined my like terms in the 0 .5 i2 and 2 .5 i2.
02:22
That becomes 3 .i2.
02:24
So we have 18 minus 3 i2 minus 6 times i1, which is i2 plus i3...