Exercise 4.3.10: Leakage put into the circuit

$Z_g\parallel Z_{\ell }=\frac{8\times 20}{8+20}\times 10^5=5.71\times 10^5{\mathrm{H}}^{-1}$, total $Z=Z_c+5.71\times 10^5=6.71\times 10^5{\mathrm{H}}^{-1}$:

$k_{\ell }={\Phi }_{\mathrm{coil}}/{\Phi }_g=1.34/0.957=1.40$ (equivalently $1+Z_g/Z_{\ell }$).

Without leakage ${\Phi }_g^{(0)}=900/(Z_c+Z_g)=1.00\mathrm{mWb}$. To get $1.00\mathrm{mWb}$ in the gap with leakage present the coil must carry $k_{\ell }\times 1.00=1.40\mathrm{mWb}$, and $NI=Z_c\times 1.40\times 10^{-3}+Z_g\times 1.00\times 10^{-3}=140+800=940\mathrm{At}$: only $40\mathrm{At}$ ($4\%$) more, because the leakage flux flows through the low-reluctance core only.

Ignoring leakage predicts $1.00\mathrm{mWb}$ in the gap. The real value is $0.96\mathrm{mWb}$, $4\%$ less, but the core carries $34\%$ more flux than predicted, which is where the error bites: the core section must be sized for ${\Phi }_{\mathrm{coil}}$, not for ${\Phi }_g$.

Node: $0.957+0.383=1.34\mathrm{mWb}$. Loop through $Z_c$ and $Z_g$: $134+766=900\mathrm{At}$. Fringing (Exercise 4.3.9) and leakage err in opposite directions on the gap flux, but leakage has a second effect fringing does not: it loads the core. A leakage coefficient of $1.1$ to $1.4$ is typical of gapped cores. It falls when the coil is placed on the gapped limb and when the window is wide compared with the gap. In transformers the same flux shows up as the leakage inductances $L_{\ell 1}$ and $L_{\ell 2}$.