00:01
So in this question we are given a circuit of which you have a very nice diagram in the question.
00:07
And for the first part, we have to calculate the currents, i1, i2, and i3, flowing in the circuit as shown on the diagram.
00:16
So i will assume that all the other letters mentioned in the figure are given.
00:23
So we know all the resistances, r1, r2 and r3.
00:26
We know the two emfs, so e1 and e2.
00:31
So the voltage is provided by the sources and we also know the internal resistance of the two sources.
00:38
So little r1 and little r2.
00:41
So in order to solve the question, we need to make use of kirchhoff's law.
00:48
So the first law of kirchhoffs tells us that the sum of the currents that enter a certain node, so sum of currents that entered a certain node, is equal to the sum of currents that go out of the node.
01:04
So for example, i will take node a as an example.
01:09
So i won't throw the entire circuit, but only the node a.
01:13
And looking at the diagram in the figure, you see that i1 goes into the node and i2 and i3 go out of the node.
01:25
So i2 goes like that and i3 goes like that.
01:29
So in this case, we can write kirkcalf's law as i1.
01:33
So with the current going in.
01:34
To the node is equal to the sum of the currents going out of the node.
01:38
So i2 plus i3.
01:41
So this is equation 1.
01:45
To write a couple of other equations that we need, we have to use kirkov's second law.
01:54
So kirkov's second law tells us that if we go around the loop, then all the voltages have to sum up to zero.
02:02
So i will, for example, draw the bottom half of the circuit just to show how kirkov's second look can be applied.
02:12
So this is node a.
02:14
In the diagram, we have a resistor here, and this is node e.
02:19
This resistor is r1.
02:22
And we also have a resistor r3 here and a voltage source on the bottom, like this.
02:34
This voltage source has an internal resistance r2 and an emfe2...