Exercise 4.3.5: Designing a small transformer

With a sinusoidal flux $\Phi ={\Phi }_{\max }\sin \omega t$ Faraday's law gives $V_{\mathrm{rms}}=4.44fN{\Phi }_{\max }=4.44fN{B}_{\max }S$, so

$N_2=N_1{V}_2/V_1=432\times 12/230=22.5$, so $N_2=23$ turns (giving $12.2\mathrm{V}$ at no load).

$Z_c=l_c/({\mu }_0{\mu }_{r}S)=0.60/(1.2566\times 10^{-6}\times 3000\times 2\times 10^{-3})=7.96\times 10^4{\mathrm{H}}^{-1}$ and

At $400\mathrm{Hz}$ the same $B_{\max }$ and $S$ need only $N_1=432\times 50/400=54$ turns. Then $L_{\mu }=54^2/Z_c=36.6\mathrm{mH}$ and $I_{\mu }=230/(2\pi \times 400\times 0.0366)=2.5\mathrm{A}$, larger, because $L_{\mu }\propto N_1^2$ fell by $64$ while $\omega$ rose by $8$. In practice one keeps more turns than the minimum, or a smaller core: at $400\mathrm{Hz}$ the core can be $8$ times smaller for the same $N_1$, which is the point of aircraft power.

Via the circuit: ${\Phi }_{\max }=B_{\max }S=2.4\mathrm{mWb}$ needs $N_1{i}_{\mu ,\max }=Z_c{\Phi }_{\max }=191\mathrm{At}$, i.e. $i_{\mu ,\max }=0.44\mathrm{A}$, rms $0.31\mathrm{A}$ as in (c). This is a few per cent of a typical full-load current for a core of this size, the price of the finite $Z_c$. One badly stacked joint, say an effective $0.1\mathrm{mm}$ gap ($Z_g=0.1\times 10^{-3}/(1.2566\times 10^{-6}\times 2\times 10^{-3})=4\times 10^4{\mathrm{H}}^{-1}$), would raise $Z$ by $50\%$ and $I_{\mu }$ to $0.47\mathrm{A}$.