Explanation

(d) When two solenoids of inductance $L_{0}$ are connected in series at large distance and current $i$ is passed through them, the total flux linkage $\phi_{total}$is the sum of the flux linkages $L_{0}i$ and $L_{0}i,\ ie,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \phi_{total} = L_{0}i + L_{0}i$
If $L$ be the equivalent inductance of the system, then
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \phi_{total} = Li$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Li = L_{0}i + L_{0}i$
or $L = 2L_{0}$
When solenoids are connected in series with one inside the other and senses of the turns coinciding, then there will be mutual inductance $L$ between them. In this case the resultant induced emf in the coils is the sum of the emf $s\ e_{1}$ and $e_{2}$ in the respective coils, $ie,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = e_{1} + e_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left( - L_{0}\frac{di}{dt} \pm L_{0}\frac{di}{dt} \right) + \left( - L_{0}\frac{di}{dt} \pm L_{0}\frac{di}{dt} \right)$
Where (+) sign is for positive coupling and (-) sign for negative coupling.
But, $e = - L.\frac{di}{dt}$
$\therefore\ \ \ \ \ \ \ \ \ - L\frac{di}{dt} = - L_{0}\frac{di}{dt} - L_{0}\frac{di}{dt} \pm 2L_{0}\frac{di}{dt}$
$ie,\ \ \ \ \ \ \ \ \ \ \ \ L = L_{0} + L_{0} + 2L_{0}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4L_{0}$ (for positive coupling)
When solenoids are connected in series with one inside the other with senses of the turns opposite, then their is negative coupling.
So, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ L = L_{0} + L_{2} - 2L_{0} = 0$