## 4 Core Modelling
### Equivalent Conductivity  
The equivalent conductivity of the laminated core was used for proper determination of the inductance. The authors decided to use equivalent electromagnetic energy stored in the region of the analysis. As a result, the field analysis may have been reduced to a one-dimensional problem, because, for a single lamination of the core, its thickness \(d\) in \(y\) direction was much smaller than its dimensions in the other directions. The model of this case is presented in a schematic manner in Figure 7.  
Figure 6: Complex permeability as a function of frequency (2D = 0.23 mm, \(\mu_{\text{r}}\) = 510, \(\gamma\) = 1.2\(\cdot 10^{6}\) S/m).  
Figure 7: A model of a laminated core with infinite \(x\), \(z\) dimensions.  
The distribution of the \(z\)-component of the magnetic intensity vector in the steel sheet of thickness \(2b\) is given by [30]:  
\[\underline{H}_{\mathrm{z}}=H_{0}\frac{\cos\mathrm{h}\alpha y}{\cos\mathrm{h} \alpha b}, \tag{8}\]  
where \(\alpha^{2}\)= j\(\omega\mu\
u_{\mathrm{e}}\) and \(\delta_{\mathrm{e}}=\sqrt{\frac{2}{\omega\mu\gamma_{\mathrm{e}}}}\).  
The energy stored in the electromagnetic field is defined by:  
\[W_{\mathrm{m}}=\frac{1}{2}\mathrm{Re}\int_{-b}^{b}\mu\cdot\underline{H}_{ \mathrm{z}}\cdot\underline{H}_{\mathrm{z}}^{*}\cdot\mathrm{d}y=\frac{1}{2} \int_{-b}^{b}\mu\cdot\left|\underline{H}_{\mathrm{z}}\right|^{2}\cdot\mathrm{d }y \tag{9}\]  
After substituting (8) into (9) and carrying out integration, the energy value is obtained, which corresponds to the equivalent conductivity \(\gamma_{\mathrm{e}}\) of the ferromagnetic block:  
\[W_{\mathrm{me}}=\frac{\mu\cdot H_{0}\cdot\delta_{\mathrm{e}}}{4}\frac{\left( \mathrm{sinh}2\left(\frac{b}{\delta_{\mathrm{e}}}\right)+\sin 2\left(\frac{b}{ \delta_{\mathrm{e}}}\right)\right)}{\left(\mathrm{cosh}\frac{b}{\delta_{ \mathrm{e}}}\cos\frac{b}{\delta_{\mathrm{e}}}\right)^{2}+\left(\mathrm{sinh} \frac{b}{\delta_{\mathrm{e}}}\sin\frac{b}{\delta_{\mathrm{e}}}\right)^{2}} \tag{10}\]  
The analogue derivation can be made for a laminated core, consisting of \(n\) sheets and thickness \(d\) and complying with relation \(2b=d\). Then the energy stored in the laminated core with conductivity \(\gamma\) takes the form of:  
\[W_{\mathrm{m}}=\frac{n\cdot\mu\cdot H_{0}\cdot\delta}{4}\frac{\left(\mathrm{ sinh}\left(\frac{d}{\delta}\right)+\mathrm{sin}\left(\frac{d}{\delta} \right)\right)}{\left(\mathrm{cosh}\frac{d}{2\delta}\cos\frac{d}{2\delta} \right)^{2}+\left(\mathrm{sinh}\frac{d}{2\delta}\sin\frac{d}{2\delta}\right)^ {2}}\ \ \mathrm{with}\ \delta=\sqrt{\frac{2}{\omega\mu\gamma}} \tag{11}\]  
The equality of the \(W_{\mathrm{me}}\) and \(W_{\mathrm{m}}\) values is accomplished when:  
\[\frac{d}{\delta}=\frac{2b}{\delta_{\mathrm{e}}}=\frac{n\cdot d}{\delta_{ \mathrm{e}}},\ \ n\cdot\delta=\delta_{\mathrm{e}},\ \ n\sqrt{\frac{2}{\omega\mu\gamma}}=\sqrt{\frac{2}{\omega\mu\gamma_{\mathrm{e}}}} \tag{12}\]  
From (12) follows the relationship for the equivalent conductivity:  
\[\gamma_{\mathrm{e}}=\frac{\gamma}{n^{2}}, \tag{13}\]  
which means that the equivalent conductivity of the laminated core decreases quadratically with the number of laminations in the transformer core.