00:01
Hello in this question to find the values of currents first let's apply kitchchof's loop law to this loop.
00:07
So applying the loop law we can write e minus i1r1 minus i1 minus i2 into r4 equals 0.
00:21
Let's substitute the values.
00:23
So 14 volt minus i1 into 1 oom minus i1 minus i 1 minus i 1 minus i 1.
00:30
2 into 2 om equals 0.
00:34
We can rewrite this equation as i1 is equal to 14 plus 2 i2 the whole divided by 3.
00:44
Let this equation be equation 1.
00:47
Now apply loop low to this loop.
00:50
So applying the loop low we can write minus i minus i 1 into r2 plus i 1 r1 plus i2 r3 equals 0.
01:06
Let's substitute the values.
01:08
So we can write minus i minus i 1 into 2 om plus i 1 into 1 om plus i 2 om plus i 2.
01:20
2 into 1 om equals 0.
01:23
We can simplify this equation as minus 2 i plus 3 i 1 plus i 2.
01:33
Equals 0.
01:36
So let this be equation 2.
01:38
Now again we have to apply the kitchops loop low to this loop.
01:42
So applying the loop low we can write minus i2 r3 minus i minus i minus i 1 plus i 2 into r5 plus i 1 minus i 2 into r4 equals 0.
02:04
Now let's substitute the values.
02:06
So we can write minus i2 into 1 om minus i -1 plus i2 into 1 -m plus i -1 -1 minus i -2 into 2 -om equals 0.
02:24
Simplifying this equation we can write minus 4 i2 -m -2 minus i plus 3 -i1 equal 0...