Two coils of self inductance L 1 and L 2 are connected in series combination having mutual inductance of the coils as M. The equivalent self inductance of the combination will be :

Question

Two coils of self inductance L 1 and L 2 are connected in series combination having mutual inductance of the coils as M. The equivalent self inductance of the combination will be : (A) (1/ L 1)+(1/ L 2)+(1/ M ) (B) L 1+ L 2+ M (C) L 1+ L 2+2 M (D) L 1+ L 2-2 M

Tardigrade Admin · Accepted Answer

Current on both the inductor is in opposite direction. Hence: L eq = L 1+ L 2-2 M

Q. Two coils of self inductance $L_{1}$ and $L_{2}$ are connected in series combination having mutual inductance of the coils as $M$ . The equivalent self inductance of the combination will be :

$\frac{1}{L _{1}} + \frac{1}{L _{2}} + \frac{1}{M}$

$L_{1} + L_{2} + M$

$L_{1} + L_{2} + 2 M$

$L_{1} + L_{2} - 2 M$

Current on both the inductor is in opposite direction.
Hence : $L_{e q} = L_{1} + L_{2} - 2 M$

Q. 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 :

L1​1​+L2​1​+M1​

L1​+L2​+M

L1​+L2​+2M

L1​+L2​−2M