00:01
Hello students, so this is the given circuit.
00:04
So we have to find the current each individual current.
00:09
So that is the a part.
00:11
So if you look at the circuit, we can see that the source current is is equal to i1 plus i4.
00:17
We can give this equation number 1 and source current is also equal to i4 plus i3 equation number 2.
00:26
So from this equation number 1 and equation number 2, we will get i1 equal to i3.
00:32
So i1 is equal to i3.
00:34
So next we have to apply kvl, kvl, kirchhoff voltage load to loop, to loop acde.
00:47
So when we apply, we will get this as 20 volt will be equal to is into r2 plus i1 into r1 plus r3.
00:56
Since the resistance are in series, current flowing will be same.
00:59
This will be equal to 20 is plus 15 i1.
01:06
We can give this is equation number 3.
01:12
So next we have to, so next we have to apply kvl, apply kvl to loop, to loop afc, sorry we have to apply kvl to loop, to loop fadc.
01:35
So we will get this as 20 volt will be equal to 20 into is plus i4 into 14 plus 6.
01:49
So those resistors r4 and r5 are in series connection.
01:54
So plus 5 i3.
01:57
So this will be equal to 20 is plus 20 i4 plus 5 i3.
02:15
So this will be equal to 20 is plus 20 into in i4, instead of i4, we can write i4 as is minus i1 plus 5 i3 and i3 is also equal to i1.
02:36
So we can substitute, we can substitute i1 where i3 is present.
02:44
So if we rearrange, we will get this equation as 20 volt equal to 40 is minus 15 i1.
02:52
So this is, this is our equation number 4.
02:54
So we can solve equation number 4 and equation number 3.
03:00
So solving equation number 4 and equation number 3, we will get like this, that's 20 is plus 15 i1 will be equal to 20 and 40 is minus 15 i1 equal to 20...