00:01
Hi, here in this given problem, current i1 is flowing through this loop, upper loop which is mentioned, which is supposed to be loop 1, the lower 1, that is a closed loop 2.
00:15
Current i1 flowing through this loop loop 1, close loop 1 in counter -clockwise direction, and current i -3 flowing through this lower loop, loop 2, close loop 2 in clockwise direction.
00:29
And this i2 that is also flowing through this lower loop in clockwise direction.
00:39
If we consider this junction to be junction a, so first of all using kirchhoff's junction rule at the junction a, i1, plus i3 will give i2, i2, which is passing through the middle branch.
01:15
Okay, now applying kirchhoff's voltage law or kirchop's loop rule in the closed loop 1.
01:38
Here, potential drop through r1, i1 into r1, that is counterclockwise, so will be taken to be negative, i2 into r2, that is also counterclockwise negative, 3 volt will be sending its current in counterclockwise direction so that is also negative.
01:57
Hence we can say minus 3 i1 algebraic sum of potential draw minus 3 i1 plus minus 4 i2 is equal to emf algebraic sum of emf only 1 emf 3 volt minus 3 volt as per sign convention so minus 3 i 1 minus 4 i2 this is i amf.
02:25
This is i .mf.
02:25
3 volt minus 3 i 2.
02:25
This is i i1 plus i 3 is equal to minus 3 expanding the bracket we can say this is minus 7 i 1 minus 4 i 3 is equal to minus 3 or simplifying it further we can say this is 7 i 1 plus 4 i 3 is equal to 3 make it equation number 1 then in closed loop 2 using same kirchof's loop rule.
03:03
And this time, i3 passing through this r2, that is clockwise.
03:10
So we'll be taken to be positive...