Two coils of inductances L1, and L2 are linked such that their mutual inductance is M

Question

Two coils of inductances L1, and L2 are linked such that their mutual inductance is M (A) L1+L2 (B) (1/2)(L1+L2) (C) (L1± L2) (D) √L1L2

Tardigrade Admin · Accepted Answer

Let us first consider a case when the total flux associated with one coil links with the other,
i.e. a case of maximum flux linkage.
Consider two coils placed adjacent to each other. Thus

M_{12} = \frac{N _{2} ϕ _{B_{2}}}{i _{1}}

and

M_{21} = \frac{N _{1} ϕ _{B_{1}}}{i _{2}}

and

L_{1} = \frac{N _{1} ϕ _{B_{1}}}{i _{1}}

and

L_{2} = \frac{N _{2} ϕ _{B_{2}}}{i _{2}}

If all the flux of coil

2

links coil

1

and vice versa, then

ϕ_{B_{2}} = ϕ_{B_{1}}

As

M_{12} = M_{21} = M

Thus we get

M_{12} M_{21} = M^{2} = \frac{N _{1} N _{2} ϕ _{B_{1}} ϕ _{B_{2}}}{i _{1} i _{2}} = L_{1} L_{2}

or

M = L_{1} L_{2}

(assuming that there is no flux leakage)
Mutual inductance is inductance of emf in a coil due to change in current in another case for two coils which are mutually coupled has mutual inductance in

= L_{1} L_{2}

This is a general result assuming that there is no flux leakage.

Q. Two coils of inductances $L_{1}$ , and $L_{2}$ are linked such that their mutual inductance is M

$L_{1} + L_{2}$

$\frac{1}{2} (L_{1} + L_{2})$

$(L_{1} \pm L_{2})$

$L_{1} L_{2}$

Q. Two coils of inductances L1​, and L2​ are linked such that their mutual inductance is M

L1​+L2​

21​(L1​+L2​)

(L1​±L2​)

L1​L2​​

Q. Two coils of inductances $L_{1}$ , and $L_{2}$ are linked such that their mutual inductance is M

$L_{1} + L_{2}$

$\frac{1}{2} (L_{1} + L_{2})$

$(L_{1} \pm L_{2})$

$L_{1} L_{2}$