## 2 Short-Circuit Current Calculation in Distribution Network with Renewable DGs  
### Traditional Simplified Short-Circuit Current Calculation  
Assuming a fault with transition resistance \(z_{f}\) occurred at node f in the distribution network, this transition resistance would not affect the node admittance matrix of the whole network. The equivalent network in the normal operation status is shown in Figure 1.  
Keeping the boundary conditions at the fault position unchanged, the original part of the network is separated from the faulty branch, as shown in Figure 1b. For a network in normal status, the occurrence of a symmetrical short-circuit fault is equivalent to the addition of an injected current, \(\dot{I}_{f}\), at the faulted node \(f\), where the short-circuit current is positive in terms of outflow from the faulted point, and the nodal current is positive in terms of injection. Therefore, the voltage at any node \(i\) in the network can be expressed as  
\[\dot{V}_{\rm i}=\sum_{j\in\rm G}Z_{\rm ii}\dot{I}_{j}-Z_{\rm ii}\dot{I}_{f} \tag{1}\]  
where symbol G represents the set of all nodes with DGs, \(\dot{V}_{\rm i}\) represents the voltage of node \(i\), and \(Z_{\rm ii}\) represents the element located at the \(i\)-th row and \(j\)-th column in the node impedance matrix, which represents the ratio of the voltage of node \(j\) to the injected current at node \(i\) if the other sources do not work. \(\dot{I}_{\rm j}\) represents the equivalent injected current at node j, and \(Z_{\rm ii}\) represents the mutual impedance between node \(i\) and the fault point \(f\).  
As shown in Equation (1), the voltage of any node \(i\) is constructed with two parts. The first part is the summary of the formulars after the symbol \(\sum\), and it represents the normal  
Figure 1: Symmetrical short-circuit fault analysis: (**a**) equivalent circuit for distribution network with a fault with transition resistance; (**b**) equivalent circuit for distribution network in normal status; (**c**) equivalent circuit for the fault branch.  
voltage of node \(i\) before the fault occurred, written as \(\dot{V}_{\rm i}^{(0)}\). The second part is the voltage at node \(i\) induced by the fault current \(\dot{I}_{\rm f}\) while all other current sources are disconnected and all other voltage sources are shorted. The summary of the two parts equals the voltage of node \(i\) after the fault occurred, which is expressed as  
\[\dot{V}_{\rm i}=\dot{V}_{\rm i}^{(0)}-Z_{\rm ff}\dot{I}_{\rm f} \tag{2}\]  
Therefore, the voltage \(\dot{V}_{f}\) of fault point \(f\) is  
\[\dot{V}_{f}=\dot{V}_{f}^{(0)}-Z_{\rm ff}\dot{I}_{\rm f} \tag{3}\]  
where \(\dot{V}_{f}^{(0)}=\sum\limits_{\rm j\in G}Z_{\rm fj}\dot{I}_{\rm j}\) represents the normal voltage of point f before the fault occurred, and \(Z_{\rm ff}\) represents the self-impedance of point \(f\), also named as the equivalent input impedance of node \(f\).  
For the fault transition resistance branch, it can be expressed as  
\[\dot{V}_{f}-z_{f}\dot{I}_{\rm f}=0 \tag{4}\]  
It can be derived from Equations (3) and (4) that  
\[\dot{I}_{\rm f}=\frac{\dot{V}_{f}^{(0)}}{Z_{\rm ff}+z_{f}} \tag{5}\]  
And then, the voltage of any node \(i\), \(\dot{V}_{i}\), can be written as  
\[\dot{V}_{i}=\dot{V}_{i}^{(0)}-\frac{Z_{\rm ff}}{Z_{\rm ff}+z_{f}}\dot{V}_{f}^{ (0)} \tag{6}\]  
The short-circuit current, \(\dot{I}_{\rm pdf}\), of any branch \(pq\), which is from node \(p\) to node \(q\) is  
\[\dot{I}_{\rm pdf}=\frac{\dot{V}_{p}-\dot{V}_{q}}{Z_{pq}} \tag{7}\]  
where \(\dot{V}_{p}\) and \(\dot{V}_{q}\) represent the voltage of node \(p\) and node \(q\), respectively, while \(Z_{pq}\) represents the impedance of branch \(pq\).