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The one-line diagram is as below,and a three-phase fault has occurred in the bus 1,now find the fault current enter image description here

\begin{align} Y=\begin{bmatrix}Y_{11}&Y_{12}\\Y_{21}&Y_{22}\end{bmatrix}=\begin{bmatrix}-j\frac{1}{0.15}-j\frac{1}{(0.1+0.1024+0.1)}&j\frac{1}{(0.1+0.1024+0.1)}\\j\frac{1}{(0.1+0.1024+0.1)}&-j\frac{1}{0.2}-j\frac{1}{(0.1+0.1024+0.1)}\end{bmatrix} \end{align}

\begin{align} Z=Y^{-1}=\begin{bmatrix} j0.1155 & j0.04598\\ j0.04598 & j0.13869\end{bmatrix} \end{align}

\$I_f=\frac{V_f}{Z_{11}}=-j8.657 p.u.\$

I have some questions about this solution, hoping someone can tell the reason

1.Why does the \$Y_{11}\$ just ignore the motor admittance,\$\frac{-j1}{0.2}\$ ?
2.\$Y_{11}\$ includes the admittance of \$Tr_1\$,\$Tr_2\$ and the generator itself,but why is \$Y_{11}=-j\frac{1}{0.15}-j\frac{1}{(0.1+0.1024+0.1)}\$,instead of \$Y_{11}=-j\frac{1}{0.15}-j\frac{1}{0.1}-j\frac{1}{0.1024}-j\frac{1}{0.1}\$ ?
3.we will calculate the fualt current \$I_f=\frac{V_f}{Z}\$,but \$X\$ is admittance,and admittance is equal to impedance when the resistance is zero,so why don't we just add all of the admittance directly ? I mean \$Z=0.15+0.1+0.1024+0.1\$,apparently it is wrong ,but why? why can't we use this method?

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