7.1 Inductors connected in series:

4.1 Motional EMF
4.2 Induced EMF in a rotating rod:

7.1 Inductors connected in series:
7.2 Inductors connected in parallel:
7.3 Mutually coupled inductors in parallel:

In this case all common current flow in all inductors because when a current $i$ flows through the first inductor (${L_1}$) it has no other way to go but pass through the second inductor and then third and so on.

Inductors are connected in series between point $A$ and $B$. The sum of the individual voltage drops across each inductor can be found using Kirchhoff’s Voltage Law (KVL).

So by Kirchhoff’s law,

$${V_T} = {V_1} + {V_2} + {V_3}$$ .........(1)

And we that the self induced emf across an inductor is given as:

$$\varepsilon = L\frac{{di}}{{dt}}$$

Or we can also write, $$V = L\frac{{di}}{{dt}}$$

So by equation $(1)$,

$${L_T}\frac{{di}}{{dt}} = {L_1}\frac{{di}}{{dt}} + {L_2}\frac{{di}}{{dt}} + {L_3}\frac{{di}}{{dt}}$$

The total inductance of the inductors connected in series can be simply found by adding together the individual inductance of the inductors. This equation holds true when there is NO mutual inductance. In this case inductors are magnetically isolated from each other.

Consider inductors $A$ and $B$ are connected in series. The self inductance of coil $A$, ${L_A}$ is and that of coil $B$ is ${L_B}$. $M$ is the mutual inductance between them. There may be two conditions.

(i) The direction of magnetic flux produce by one is in same direction of other. We know that the flux of coil $B$ is linked with coil $A$ and it is in same direction with the flux produce by coil $A$, itself. Hence, total effective inductance of coil $A$ is ${L_A}$+M.

At the same time, the flux of coil $A$ is linked with coil $B$ and it is in same direction with the flux produce by coil $B$, itself. Hence, total effective inductance of coil $B$ is ${L_A}$+M.

Total effective inductance of the series connected inductors $A$ and $B$ will be,

$${L_A} + {L_B} + 2M$$

(ii) The direction of magnetic flux produce by one is in opposite direction of other. We know that the flux of coil $B$ is linked with coil $A$ and it is in opposite direction with the flux produce by coil $A$, itself. Hence, total effective inductance of coil $A$ is ${L_A}$ - M.

At the same time, the flux of coil $A$ is linked with coil $B$ and it is in opposite direction with the flux produce by coil $B$, itself. Hence, total effective inductance of coil $B$ is ${L_B}$ - M.

Total effective inductance of the series connected inductors $A$ and $B$ will be,

$${L_A} + {L_B} - 2M$$