00:01
Hi, so we are given that a circuit so we have to find the magnitude of voltage vab is equals to question mark.
00:13
So, first of all let's have a look to the circuitry.
00:16
So this is what the circuitry which we have been given here.
00:20
So according to question r1 resistance is of 1 .4 ohm, r2 is equals to 2 ohm, r3 is equals to sorry not r3, epsilon 1 emf is equals to 2 volt, epsilon 2 is equals to 6 volt and epsilon 3 means emf 3 is also equals to 6 volt.
00:47
So we just have to find the voltage drop across this vab point.
00:50
So let it started with solution.
00:54
So the concept which we are going to use for this question will be the kirchhoff's voltage law that we called as kvl.
01:04
The summation of voltage inside a closed loop is equals to 0.
01:08
So by using this if you apply the loop analysis.
01:13
So first of all let's see the current flowing throughout it is i1 from this is i2 and from this branch will be i1 plus i2.
01:24
So now i am considering this as loop first, this as loop second.
01:30
So applying the kvl to the first loop we get an equation e1 minus of i1 r1 plus i2 r2 this should be equals to minus of epsilon 2 minus of i1 r1 this is equals to 0.
01:53
So just need to substitute the values epsilon 1 is 2 minus i1 known to us then r1 is 1 .4 plus i2 unknown to us r2 is 2 ohms minus of epsilon 2 is 6 then minus of i1 will be as it is 1 .4 is equals to 0.
02:15
So on combining you can clearly see this two points are equal minus 6 plus 2 will become minus 4.
02:23
So we get negative of 2 .8 i1 plus 2 times of i2 negative of 4 is equals to 0 or rather we can say 2 .8 i1 minus of 2 i2 plus of 4 is equals to 0.
02:40
Let's see equation first.
02:43
Now in similar way we can apply the kvl to the second loop...