00:01
Hi, this given problem that is based upon kirchop's junction rule and kirchov's voltage rule means kirchop's current law and kirchop's voltage law.
00:13
So before starting this problem, we name this loop as a, b, c, d and e and f.
00:25
The values of resistors are 1.
00:29
It is 6 .5.
00:33
8 -oom, r2 that is 23 .8 om, r3 70 .0 om.
00:44
Suppose current i -1 comes out of the battery of mfe2.
00:51
This emfe2 that is given as 366 volt and e1 that is 40 volt.
01:07
Current i2 comes out of e2.
01:11
Then this current will be divided into two parts at this junction.
01:17
At junction e, current, this current that is i1, i1 coming out of the battery of mfe2.
01:30
I2 will be passing through the battery of emfe1 and remaining current i3 will pass.
01:39
Towards will move towards right so using kirchov's current law at the junction e this current i -1 which is entering to the junction e will be equal to algebraic sum of i2 and i3 we consider it to the equation number one now using and that is for the first part of the which we are doing now using kirchhoff's voltage lock in the closed loop a b e f a a here in this loop current i1 passing through r2 in clockwise direction so potential drop will be taken to be positive i one passing through this 23 .8 ome and i2 passing through this r1 means a t6 .8 ome.
02:56
I2 is passing through a t6 .8 om that is also clockwise so positive.
03:00
Mf e .e1 sending its current clockwise so positive.
03:04
E2 also positive as sending its current in clockwise direction.
03:09
Now kirchchov's voltage law it says that algebraic sum of potential drops.
03:14
So starting with this 23 .8 multiplied by i.
03:23
Plus 86 .8 multiplied by i 2 is equal to 366 plus 40.
03:33
Algebra sum of potential drops is equal to algebraic sum of enfs.
03:39
Or simplifying means dividing both the sides by 2 we get 11 .9i1 plus 43 .4 i .2.
03:51
That is equal to 203.
03:54
Equation number 2.
03:59
Now using same kirchhoff's voltage law in the closed loop b, c, d, e, b and this time potential drop through 23 .8 om for this loop that is counterclockwise.
04:24
So negative.
04:25
I3 passing through r3 that is also counterclockwise.
04:28
So negative...