00:01
Hello students, in this portion we have given four inductors l1, l2, l3 and l4.
00:08
Also we have given the mutual inductance between coil 2 and 5 as well as between 4 and 5 and the value of this is 6 .2.
00:20
Also we have given the potential difference between points a and b so va -vb that is v equal to v max e to the power minus t by tau.
00:43
Now for self -inductive for self -induction l1 and l2 are in series so we can write l will be equal to l1 plus l2.
01:04
We have the values 2 .8 plus 8 .3 that is equal to 11 .1.
01:14
Now there is also mutual inductance given between the two coils so there is a corresponding emf also due to mutual inductance.
01:38
Now applying the same we can write va -l1 plus l2 di1 by dt minus m di by dt minus vb that is equal to 0.
01:59
So we can write this difference of potential va -vb that is equal to l1 plus l2 plus m di1 on dt.
02:15
Now substituting the values here so 11 .1 plus 6 .2 di1 on dt.
02:28
So for the solving we can write 17 .3 di1 by dt.
02:33
On left hand side we have potential difference va -vb.
02:38
Now we have given the coefficient difference potential difference between a and b is given as v max e to the power minus t by tau...