# Efficient Zero-Sequence Impedance Measurement in Autotransformers Using Low-Voltage Excitation
## 3 YNa0d11 Three-Winding Autotransformer Zero-Sequence Impedance Field Test Method  
### One-Phase Disconnection Test  
The connection between the high- and medium-voltage sides of the YNa0d11 auto-transformer is star grounding, with the low-voltage side being the triangle connection. To measure the zero-sequence equivalent impedance of individual windings, the following 3 disconnection tests are required on the low-voltage side:  
1. [label=(0)]
2. Test 1: Positive-sequence excitation on the low-voltage side; while one phase of the high-voltage winding is disconnected, the other two phases are short circuited to earth and the medium-voltage winding is completely disconnected. 3. Test 2: Positive-sequence excitation on the low-voltage side; while one phase of the medium-voltage winding is disconnected, the other two phases are short circuited to earth and the high-voltage winding is completely disconnected. 4. Test 3: Positive-sequence excitation on the low-voltage side; while one phase of the medium- and the high-voltage winding is disconnected, the other two phases are short circuited to earth. Tests 1 and 2 have similar test content and wiring methods, measuring the zero-sequence impedance to ground of the high-voltage side \(Z_{\text{Hg-0}}\) and the medium-voltage side \(Z_{\text{Tg-0}}\), respectively. Here, \(Z_{\text{Hg-0}}=Z_{\text{H-0}}=Z_{\text{H-0}}+Z_{\text{g-0}}\), \(Z_{\text{Tg-0}}=Z_{\text{T-0}}+Z_{\text{g-0}}\). The connection for Test 1 is illustrated in Figure 9. The subscripts \"a, b, c\" in Figure 9 represent the parameters on the low-voltage side of the transformer. The zero-sequence impedance of this autotransformer can be derived from the composite sequence network diagram in Figure 10 as \(Z_{\text{Hg-0}}=(V_{1}-I_{1}\,Z_{\text{HL-1}})/I_{0}\). Considering the phase variations introduced by different connection methods, the Yd-11 coupling method, more commonly used in power systems, is taken as an example. In this method, the positive-sequence component of the voltage and current on the low-voltage side of the auto-transformer leads the high- and medium-voltage side by \(30^{\circ}\), while the negative-sequence component lags the high- and medium-voltage side by \(30^{\circ}\). The phase relationship between current and voltage is shown in Figure 11. Therefore, it can be found that:  
\[\left\{ \begin{array}{l}\dot{I}_{a}=\frac{\sqrt{3}}{2}(I_{1}+I_{2})+j\frac{1 }{2}(I_{1}-I_{2})\\ \dot{I}_{b}=-j(I_{1}-I_{2})\\ \dot{I}_{c}=-\frac{\sqrt{3}}{2}(I_{1}+I_{2})+j\frac{1}{2}(I_{1}-I_{2})\end{array} \right. \tag{13}\]  
Figure 9: Test 1 wiring diagram. Based on Equation (12), \(\dot{l}_{a}\) and \(\dot{l}_{c}\) have the same imaginary part, and opposite real parts. By assuming that \(\left|\dot{l}_{a}\right|=\left|\dot{l}_{c}\right|=I\) and the angle between \(\dot{l}_{a}\) and \(\dot{l}_{1}\) is \(\theta\), it can be found that:  
\[\left\{ \begin{array}{l}\theta=\cos^{-1}\biggl{(}\frac{\left|\dot{l}_{b} \right|}{2I}\biggr{)}\\ \dot{l}_{1}=\frac{2}{\sqrt{3}}I\sin(\theta+30^{\circ})\\ \dot{l}_{2}=\frac{2}{\sqrt{3}}I\sin(\theta-30^{\circ})\end{array} \right. \tag{14}\]  
By assuming that \(\left|\dot{V}_{ab}\right|=\left|\dot{V}_{bc}\right|=V\) and combining this with Equation (5), it can be found that:  
\[\left\{ \begin{array}{l}\varphi=\cos^{-1}\bigl{(}\frac{\left|\dot{V}_{a} \right|}{2V}\bigr{)}\\ V_{1}=-\frac{2}{\sqrt{3}}V\cos(\varphi-30^{\circ})\\ V_{2}=-\frac{2}{\sqrt{3}}V\cos(\varphi+30^{\circ})\end{array} \right. \tag{15}\]  
By combining Equations (14) and (15), the zero-sequence impedance \(Z_{\text{Hg-0}}\) can be found, as well as \(Z_{\text{Tg-0}}\). The test connection for Test 3 is illustrated in Figure 12. The subscripts \"a, b, c\" in Figure 12 represent the parameters on the low-voltage side of the transformer.