Exercise 4.2.10: Energy of two coupled coils

$L_{12}=k\sqrt{L_{11}{L}_{22}}=0.6\sqrt{10\times 40}\mathrm{mH}=12\mathrm{mH}$.

$\partial W_m/\partial I_2=L_{12}{I}_1+L_{22}{I}_2=0$ gives $I_2=-L_{12}{I}_1/L_{22}=-0.6\mathrm{A}$, and

$W_{\min }>0$ whenever $k<1$: the energy expression is positive definite exactly when $L_{12}^2<L_{11}{L}_{22}$, which is the proof that $k\le 1$ given in Application 4.2.2. The opposing current in coil 2 cancels most of coil 1's flux through coil 2 but cannot cancel the leakage.

In series the same current flows, $I_1=\pm I_2=I$, and $W_m=\frac{1}{2}\left(L_{11}+L_{22}\pm 2L_{12}\right)I^2$:

With $I_2=-0.6\mathrm{A}$ in (b) the formula gives $20-14.4+7.2=12.8\mathrm{mJ}$, the minimum found in (c). Measuring $L_{\text{aid}}$ and $L_{\text{opp}}$ with an inductance meter is the practical way of finding $L_{12}=(L_{\text{aid}}-L_{\text{opp}})/4=(74-26)/4=12\mathrm{mH}$. With $k\to 1$ (ideal transformer) $W_{\min }\to 0$: a short-circuited secondary then stores no energy, and $L_{\text{opp}}\to (\sqrt{L_{11}}-\sqrt{L_{22}}{)}^2$.