Berechnung am Beispiel aus Abb. 3 :
{\large\begin{array}{l}geg.:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,ges.:\,{{I}_{1}}\\{{U}_{1}}=9\,\text{V}\\{{N}_{1}}=300\,\text{Wdgn};\,\,\,{{N}_{2}}=900\,\text{Wdgn}\\R=\,500\,\text{ }\!\!\Omega\!\!\text{ }\,\,\end{array} }
Lösung:
{\large\begin{array}{l}\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{{{N}_{1}}}{{{N}_{2}}}\\{{U}_{2}}\,=\,{{U}_{1}}\cdot \frac{{{N}_{2}}}{{{N}_{1}}}\,=\,9\,\text{V}\,\cdot \,\frac{900\,\text{Wdgn}}{300\,\text{Wdgn}}\,=\,27\,\text{V}\\\\{{I}_{2}}=\frac{{{U}_{2}}}{R}\,=\,\frac{27\,\text{V}}{500\,\Omega }\\{{I}_{2}}\,=\,54\,\text{mA}\\\\\frac{{{I}_{1}}}{{{I}_{2}}}\,=\,\frac{{{N}_{2}}}{{{N}_{1}}}\\{{I}_{1}}=\,{{I}_{2}}\,\cdot \,\frac{{{N}_{2}}}{{{N}_{1}}}\,=\,54\,\text{mA}\cdot \frac{\text{900}\,\text{Wdgn}}{\text{300}\,\text{Wdgn}}\\{{I}_{1}}\,=\,162\,\text{mA}\end{array} }