In Fig. coil `1` and coil `2` are wound on a long cylindrical insulator. The ends `A'` and `B` are joined together and current `I` is passed. Self-inductance of the two coils are `L_(1)` and `L_(2)`, and their mutual inductance is `M`.
a. Show that this combination can be replaced by a single coil of equivalent inductanCe given by
`L_(eq) = L_(1) + L_(2) + 2M`.
b. How could the coils be reconnected by yieldings an equivalent inductance of `L_(eq) = L_(1) + L_(2) - 2M`.

Explanation

The emf induced across coil BB. must be
`V_(B)=-L_(B)(di_(B))/(dt)+M(di_(A))/(dt)`
Note that the magnetic field due to the coil itself and due to coil AA. are in opposite directions. So, their induced emf will also have opposite signs. Here M is the (magnitude of the) mutual inductance between the coils. Since the coils are connected in series `i_(A) = i_(B) = i`, so that
`V_(B)=-(L_(B)-M)(di)/(dt)`
Similarly, we find that the emf induced across coil AA.is
`V_(A)=-(L_(A)-M)(di)/(dt)`
The emf across the entire series coil assembly is the sum of `V_(A)` and `V_(B)`:
`V=V_(A)+V_(B)=-(L_(A)+L_(B)-2M)(di)/(dt)`