00:01
Hello students, from the circuit we can see that current i1 is equal to sum of current i2 and i3.
00:09
Now using kirchhoff voltage law in loop 1, we can write i1 into r1 plus r2 plus r4 into i2 is equal to v1.
00:25
Now substituting the values here, we can write that is i1 is replaced with i2 plus i3 into r1 plus r2 plus r4 into i2 is equal to v1.
00:47
As we are having i2 r1 is also here, so splitting that we can write i3 r1 plus r1 plus r2 plus r4 and i2 is common for these three taking outside is equal to v1.
01:03
Now substitute the values, we need to calculate i3 r1 36 ohm plus r1 36 plus r2 32 plus r4 36 which is given i2 as it is v1 is 112 volt.
01:24
So from this 36 i3 plus 104 i2 is equal to 112, consider this as equation 1.
01:39
Now for loop 2, i1 r1 plus r3 r5 i3 is equal to v2.
01:54
Again considering i1 as i2 plus i3, we can write i2 r1 plus taking here r1 plus r3 plus r5 and i3 is common for these three is equal to v2.
02:09
Now substitute the values i2 as it is r1 36 r1 36 plus r3 42 plus r5 39 i3 as it is is equal to v2.
02:27
Now from this calculation we get 36 i2 plus 117 i3 is equal to v2 is 118.
02:43
So now consider this as equation 2...