Two coils having the same self-inductance are connected in series. When a current $\mathrm{I}$ flows through the coils, the magnetic energy stored in their fields is W joules. If the connections of one coil are interchanged and the current is reduced to $(1/2)\mathrm{I}$ , the energy stored is again W joules. Calculate the ratio of the mutual inductance and the self-inductance.

The effective inductances of the circuit, for the two connections are :
$\begin{array}{c}{\mathrm{L}}_{\mathrm{A}}=\mathrm{L}+\mathrm{L}-2\mathrm{M} =2\mathrm{L}-2\mathrm{M}\\ \mathrm{and}, {\mathrm{L}}_{\mathrm{B}}=\mathrm{L}+\mathrm{L}+2\mathrm{M}=2\mathrm{L}+2\mathrm{M}\\ \end{array}$ 
  Stored energy in the circuit 1,   
and the stored energy in the circuit 2,  
Now, we are given that 

$\begin{array}{c}{\mathrm{I}}_2=\frac{1}{2}{\mathrm{I}}_1=\frac{1}{2}\mathrm{I}\text{ and }{\mathrm{W}}_{\mathrm{A}}={\mathrm{W}}_{\mathrm{B}}\\ \therefore \frac{1}{2}{\mathrm{L}}_{\mathrm{A}}{\mathrm{I}}^2={\mathrm{W}}_{\mathrm{A}}=\mathrm{W}={\mathrm{W}}_{\mathrm{B}}=\frac{1}{2}{\mathrm{L}}_{\mathrm{B}}{\left(\frac{1}{2}\mathrm{I}\right)}^2\\ \therefore \frac{1}{2}{\mathrm{L}}_{\mathrm{A}}=\frac{1}{8}{\mathrm{L}}_{\mathrm{B}}\\ {\mathrm{L}}_{\mathrm{A}}=\frac{1}{4}{\mathrm{L}}_{\mathrm{B}}\Rightarrow 2(\mathrm{L}-\mathrm{M})=\left(\frac{1}{4}\right)2(\mathrm{L}+\mathrm{M})\\ \therefore 4(\mathrm{L}-\mathrm{M})=\mathrm{L}+\mathrm{M}\text{ ⇒ }3\mathrm{L}=5\mathrm{M}\\ \therefore \frac{\mathrm{M}}{\mathrm{L}}=\frac{3}{5}=0.6\end{array}$