00:01
In this problem, we have been given the following circuit and the values of the cells emf and the internal resistances are given.
00:10
And both these internal resistances are equivalent to 2 -on and the external resistance are here that's connected through this point d.
00:18
That's 30 -on.
00:20
And we have to determine the current that's observed through e1.
00:24
So let's say we take this current as i -1 and the current that's observed through the cell e2.
00:31
So let's take that as i2.
00:34
And also we have to figure out the current that's flowing through this resistor r.
00:40
So let's take that current as i are which we have to determine.
00:46
So first applying kirchav's current law, we observed that at this point a, the total current which is coming to this junction, that's i1 and i2.
00:57
And the total current which will be leaving this junction, that will be i1 plus i2 and it will be same so we can see ir will be equal to i1 plus i2 so this is in accordance with kirchhoff's current law and now let's make use of kirchof's voltage law so here we will be considering the loops so first let's consider this loop let's take this loop as loop two and we take this as loop three so in loop two we observe that using kirchhoff's voltage law which we also call it as kvl that's just standing for kitchf's voltage law so we start at any point let's say we start at point c so the potential at this point c let's consider that as zero volt and respect to this the potential at this point that will be 24 volt because even is having 24 volt and there will be potential drop across this resistor r1 because we know that in the direction of current the potential decreases.
02:00
So we get the potential at this point which is equal to the potential at a and that's 24 minus i1 r1.
02:10
R1 is 2.
02:11
Let's substitute that.
02:13
And here we observe potential at this point.
02:17
That will be the potential at a plus i into r2 because there will be potential gain which can be just figured out using omstop.
02:28
So it will be 2 times i2 and the potential at b will be the potential which we figured out minus the potential difference of the cell 2 and we observe that potential at b and c that's same so this equated to 0 here so from this kirchhoff's voltage law we're going to get one equation and here we observe that this is 24 so first let's divide throughout by 2 so we get 12 minus i1 1.
03:00
Plus i2 minus 20 and that's equal to 0.
03:04
So we get i2 minus i1.
03:07
That will be eight.
03:09
So let's mark it as equation one here.
03:13
And now we use same kirchhoff's voltage law to this loop three.
03:16
So using again kirchhoff's voltage lock in loop three.
03:22
So here let's move in the clockwise direction as indicated here.
03:27
So first let's start with this point b and the potential at this.
03:31
Point that will be plus 40 volt because the potential difference of the cell is 40 volt and now we can figure out the potential at a so we are moving in the direction of current here so potential will be dropped so it will be minus 2 i2 and now we observe again in the direction of the current we are moving and the potential will be dropped across this resistor r so we just subtract that to get the potential at d and that will be minus 30 into the current through this.
04:03
And the current can be replaced with i1 and i2...
