Two coils of self inductance L₁ and L₂ are connected in series combination having mutual inductance of the coils as M. The equivalent self inductance of the combination will be :

Solution

Physics sir · 4 steps

Follow the current through each coil: it runs opposite ways, so the mutual flux subtracts.

1. Read the figure. The current I enters the left coil at its left end and flows to the right. It then goes along the bottom wire, enters the right coil at its right end and flows to the left. The two coils share the same axis and are wound the same way, so their magnetic fields point in opposite directions.
2. Flux in each coil. Coil 1 carries its own flux L₁ I. The opposing field of coil 2 removes MI from it. Coil 2 likewise has L₂ I - MI.

   Φₜₒₜₐₗ = (L₁ I - MI) + (L₂ I - MI) = (L₁ + L₂ - 2M) I

3. Why 2M. Mutual inductance acts both ways: coil 2 changes the flux in coil 1, and coil 1 changes the flux in coil 2. That is why M appears twice.
4. So. Since Φₜₒₜₐₗ = L_(eq) I, the equivalent inductance is L₁ + L₂ - 2M, option (d).

The idea to keep. For coils in series, check whether the currents make their fields aid or oppose. Aiding gives L₁ + L₂ + 2M, opposing gives L₁ + L₂ - 2M. The 2 comes from each coil acting on the other.

Answer from the NTA final answer key. Solution by PrepWiser.