00:01
In this problem, we have been given this circuit and we have to write down the loop as well as the junction equations.
00:09
And we will do that using kirchav's two laws.
00:13
So one is the kirchhoff's current law which will apply at different junctions according to which the total current which is coming to the junction that's equal to the total current which is leaving out of that junction.
00:26
And we will figure out another equation using kirchhoff's voltage law, which is also called as the loop law.
00:34
And in this we are going to consider a circuit and in that the total potential that will be dropped in a loop that will be equal to zero.
00:44
So first we observe that there are two junctions here.
00:47
One is, let's say this is junction j1 and this is j2.
00:52
And let's consider the current which is flowing in different arms.
00:58
Let's say this is i1 and let's say this is current i2.
01:03
So if we observe at junction j1, at this point, the total current which is coming, that's i1 plus i2, and that will be equal to the total current which is leaving after this junction.
01:16
So that's i3, let's say.
01:18
So i1 plus i2, that will be i3.
01:21
And similarly, if we observe the another junction that's j2 here as indicated in the diagram, here the total current that's coming to this junction, that's i3, and that will be equal to the total current that's leaving of this junction.
01:38
So here as this i1 was coming out of this battery with voltage v -0 -01, so same current will be going into this battery as well, and same thing applies to the right arm here.
01:48
So we get i3 as i1 plus i2 and we see that same equation we have derived that we got for the junction j1.
01:58
And now we just see that there are two loops basically here.
02:03
So let's consider this as one loop and considering this as another loop.
02:08
Let's name that as loop 1 and loop 2.
02:11
So we will apply kirchap's voltage law in loop 1 and we will use oms law to compute the potential drop across each resistor given by the equation v equals i times r.
02:24
So let's say we start with this point and the potential is zero volt.
02:29
So the potential here, it will be v -0 -0 -1.
02:32
So we are moving from this point with zero potential and we add v -001.
02:39
And there will be potential drop across this resistor r -001...