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 :

Key Concepts Needed

1. Self-Inductance ($L$)
   • For one coil, $L= \dfrac{N\Phi}{I}$, where $\Phi$ is self-flux.
2. Mutual Inductance ($M$)
   • When current $I$ in coil-1 produces flux $\Phi_{21}$ linking coil-2, we define $M_{21}= \dfrac{N_2\Phi_{21}}{I_1}$ (and similarly for the reverse process).
3. Series Connection of Inductors
   • Same current $I$ flows through both coils.
   • The total magnetic flux linking each coil has two parts: self-flux and mutual flux.

Building The Equivalent Inductance

Flux in Coil-1 $\Phi_1= L_1 I \;\pm\; M I$ Flux in Coil-2 $\Phi_2= L_2 I \;\pm\; M I$ (The sign depends on whether mutual flux aids or opposes self-flux.)

Total flux linkage for the series pair: $N_1\Phi_1 + N_2\Phi_2 = (L_1 + L_2 \;\pm\; 2M)\,I$

Equivalent inductance $L_{eq}$ is defined by $\text{(total linkage)} = L_{eq} I$

Hence $L_{eq}=L_1 + L_2 \;\pm\; 2M$

Why each step?

1. We start from definition of inductance (linkage per current).
2. In series, same current simplifies algebra.
3. Mutual flux adds twice (once in each coil), giving $2M$.
4. Sign chosen by magnetic orientation of coils.

Imagine Two Springs Making One Bigger Spring

Think of two toy springs (the coils).

1. Each spring alone has its own stiffness — we call it $L_1$ or $L_2$ (self-inductance).
2. Close together, when one spring moves, it tugs the other a bit. That tugging power is called mutual inductance $M$.
3. If you tie the springs end-to-end (series), the total stiffness depends on whether the tugs help or fight each other.
   • If both springs twist the same way (they help), the stiffness adds extra: $+2M$.
   • If they twist opposite ways (they fight), you lose some stiffness: $-2M$.

So the total spring (equivalent inductance) is:

$L_{eq}=L_1+L_2\;\pm\;2M$

Choose + when the coils aid each other, – when they oppose.

Step-by-Step Calculation

1. Write flux linkages for each coil:

$\lambda_1 = L_1 I \;\pm\; M I$ $\lambda_2 = L_2 I \;\pm\; M I$

2. Add them (series means same current, total linkage is sum):

$\lambda_{total}=\lambda_1+\lambda_2 = \left(L_1 + L_2 \;\pm\; 2M\right) I$

3. Define equivalent inductance $L_{eq}$ by

$\lambda_{total}=L_{eq}\,I$

4. Compare and read off:

$\boxed{\displaystyle L_{eq} = L_1 + L_2 \;\pm\; 2M}$

Take +2M for series-aiding, −2M for series-opposing.