00:01
Hello students in the question we are given a circuit that is assumed current i1 flows in this loop and current i2 flows in this loop.
00:12
So by kirchhoff's current law we would have a junction a summation of currents would be equal to zero.
00:20
So here we can assume current flows in this direction and current flowing would be i2 minus i1.
00:27
We can assume direction of loops like this.
00:33
So from kirchhoff's second law or voltage law we can write summation change in potential it is equal to summation of current into resistance is equal to zero.
01:00
Therefore we can write in closed loop abcd here we have we are moving from negative terminal to positive terminal so it is positive 58 volt here the current flow and the direction of loop is same so we would have minus i1 into 120 again the direction of current flow and direction of loop it is same.
01:35
So here we would have minus i1 82 now here direction of current it is opposite to direction of loop here we can write plus i2 minus i1 into 64 is equal to zero.
01:57
So here we can write this is equal to 58 minus i1 120 minus i1 82 plus i2 64 minus i1 64 is equal to zero or we get from here 58 minus i1 120 plus i1 120 minus 266 i1 plus 64 i2 it is equal to zero or we can write 266 i1 minus 64 i2 is equal to 58.
02:30
This is equation one.
02:34
Similarly in another loop we can write minus i2 25 minus i2 minus i1 64 plus i2 into 110 plus i2 into 120.
02:51
So here we can write minus i2 25 minus i2 64 plus i1 into 64 plus i2 into 110 plus 3 .0 is equal to zero or we get minus 199 i2 plus 64 i1 plus 3 it is equal to zero or we can write from here i2 it is equal to 3 plus 64 i1 divided by 199.
03:31
Here we can substitute this value of i2 in equation one...