## 3 Development of Scott Transformer with Electromagnetic Transients Program
### Electromagnetic Transients Modeling
#### 3.1.2 Current Injection Model for Scott Transformer  
Considering the n-winding transformer in (5), the voltage and current input matrices are defined as follows:  
\[[\mathbf{v}(t)]=\left[ \begin{array}{cccc}v_{1}(t)&v_{2}(t)&\cdots&v_{n}(t) \\ \mathrm{[}i(t)]=\left[\begin{array}{cccc}i_{1}(t)&i_{2}(t)&\cdots&i_{n}(t) \end{array}\right]^{\mathrm{T}}\end{array} \right. \tag{6}\]  
Subsequently, to include the matrices \(\mathbf{R}\) and \(\mathbf{L}\), the voltage and current are defined as Equation (7) based on Equation (5). \[[\mathbf{v}(t)]=[\mathbf{R}][\mathbf{i}(t)]+[\mathbf{L}]\frac{d}{dt}[\mathbf{ i}(t)] \tag{7}\]  
Adjusting these equations to EMT programs, the matrix representation of the transformer can be represented by being numerically integrated over a time step using trapezoidal integration. Then, the voltage and current in Equation (7) can be reorganized as:  
\[\frac{[\mathbf{v}(t)]+[\mathbf{v}(t-\Delta t)]}{2}\Delta t=[\mathbf{R}]\frac{ [\mathbf{i}(t)]+[\mathbf{i}(t-\Delta t)]}{2}\Delta t+[\mathbf{L}][[\mathbf{i}( t)]-[\mathbf{i}(t-\Delta t)]] \tag{8}\]  
Figure 6: Configuration of an equivalent circuit for a single-phase transformer. It can be represented in the form of the current injection model as follows [14]:  
\[\left[\left[\mathbf{R}\right]\frac{\Delta\mathrm{t}}{2}+\left[\mathbf{L}\right] \right]\left[\mathbf{i}(\mathbf{t})\right]+\left[\left[\mathbf{R}\right]\frac{ \Delta\mathrm{t}}{2}-\left[\mathbf{L}\right]\right]\left[\mathbf{i}(\mathrm{t }-\Delta\mathrm{t})\right]=\frac{\Delta\mathrm{t}}{2}\left[\mathbf{v}(\mathrm{ t})\right]-\frac{\Delta\mathrm{t}}{2}\left[\mathbf{v}(\mathrm{t}-\Delta \mathrm{t})\right] \tag{9}\]  
Equation (9) can be rearranged to express it as a conductance matrix \(\mathbf{G}\):  
\[ \begin{array}{c}\left[\mathbf{i}(\mathrm{t})\right]=\left[\left[\mathbf{R} \right]\frac{\Delta\mathrm{t}}{2}+\left[\mathbf{L}\right]\right]^{-1}\!\! \frac{\Delta\mathrm{t}}{2}\!\left[\mathbf{v}(\mathrm{t})\right]-\left[\left[ \mathbf{R}\right]\frac{\Delta\mathrm{t}}{2}+\left[\mathbf{L}\right]\right]^{- 1}\!\!\frac{\Delta\mathrm{t}}{2}\!\left[\mathbf{v}(\mathrm{t}-\Delta\mathrm{t })\right]\\ \qquad\qquad-\left[\left[\mathbf{R}\right]\frac{\Delta\mathrm{t}}{2}+\left[ \mathbf{L}\right]\right]^{-1}\!\left[\left[\mathbf{R}\right]\frac{\Delta \mathrm{t}}{2}-\left[\mathbf{L}\right]\right]\!\left[\mathbf{i}(\mathrm{t}- \Delta\mathrm{t})\right]\end{array}  \tag{10}\]  
Consequently, it can be formed as follows:  
\[\left[\mathbf{i}(\mathrm{t})\right]=\left[\mathbf{G}\right]\!\left[\mathbf{v}( \mathrm{t})\right]+\left[\mathbf{i}_{\mathbf{h}}(\mathrm{t})\right] \tag{11}\]  
where \(\left[\mathbf{G}\right]=\left[\left[\mathbf{R}\right]\frac{\Delta\mathrm{t}}{ 2}+\left[\mathbf{L}\right]\right]^{-1}\!\!\frac{\Delta\mathrm{t}}{2}\), \(\left[\mathbf{i}_{\mathbf{h}}(\mathrm{t})\right]=\left[\mathbf{G}\right]\! \left[\left[\mathbf{v}(\mathrm{t}-\Delta\mathrm{t})\right]-\left[\left[ \mathbf{R}\right]-\left[\mathbf{L}\right]\frac{2}{\Delta\mathrm{t}}\right] \!\left[\mathbf{i}(\mathrm{t}-\Delta\mathrm{t})\right]\right]\)\(\mathbf{G}\) is composed of \(\mathbf{R}\) and \(\mathbf{L}\) matrices, and \(\mathbf{i}_{\mathbf{h}}(\mathrm{t})\) indicates the current of the previous time step. Meanwhile, the current injection method including the Dommed algorithm is used in the EMT program to adjust the formula for each solution time step [14]. Unlike the typical Norton equivalent circuit in which one end is connected to the ground, the current injection model handles the voltage of each terminal rather than measuring the voltage across the terminals of each winding [31]. Based on the numerical trapezoidal method including Equations (8)-(11), the current injection model can be formed using the voltage and current at both ends as follows:  
\[\left[ \begin{array}{c}i_{1}^{+}(\mathrm{t})\\ i_{2}^{+}(\mathrm{t})\end{array} \right]=\left[\mathbf{G}_{\mathrm{TR}}\right] \left[ \begin{array}{c}v_{1}^{+}(\mathrm{t})-v_{1}^{-}(\mathrm{t})\\ v_{2}^{+}(\mathrm{t})-v_{2}^{-}(\mathrm{t})\end{array}