# Unbalanced, Three phase, Three wire, Y Connected Load

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For the system shown in Fig. 1, the required equations can be derived by first applying Kirchhoff’s voltage law around each closed loop to produce
Fig. 1: Unbalanced, three-phase, three-wire, Y-connected load
$$\begin{array}{l}\mathbf{E}_{A B}-\mathbf{V}_{a n}+\mathbf{V}_{b n}=0 \\\mathbf{E}_{B C}-\mathbf{V}_{b n}+\mathbf{V}_{c n}=0 \\\mathbf{E}_{C A}-\mathbf{V}_{c n}+\mathbf{V}_{a n}=0\end{array}$$
Substituting, we have
$$\mathbf{V}_{a n}=\mathbf{I}_{a n} \mathbf{Z}_{1} \quad \mathbf{V}_{b n}=\mathbf{I}_{b n} \mathbf{Z}_{2} \quad \mathbf{V}_{c n}=\mathbf{I}_{c n} \mathbf{Z}_{3}$$
$$\begin{array}{l} \bbox[10px,border:1px solid grey]{\mathbf{E}_{A B}=\mathbf{I}_{a n} \mathbf{Z}_{1}-\mathbf{I}_{b n} \mathbf{Z}_{2}} \quad \text{(1)}\\ \bbox[10px,border:1px solid grey]{\mathbf{E}_{B C}=\mathbf{I}_{b n} \mathbf{Z}_{2}-\mathbf{I}_{c n} \mathbf{Z}_{3}} \quad \text{(2)}\\ \bbox[10px,border:1px solid grey]{\mathbf{E}_{C A}=\mathbf{I}_{c n} \mathbf{Z}_{3}-\mathbf{I}_{a n} \mathbf{Z}_{1}} \quad \text{(3)} \end{array}$$
Applying Kirchhoff's current law at node $n$ results in
$$\mathbf{I}_{a n}+\mathbf{I}_{b n}+\mathbf{I}_{c n}=0 \quad \text { and } \quad \mathbf{I}_{b n}=-\mathbf{I}_{a n}-\mathbf{I}_{c n}$$
Substituting for $\mathbf{I}_{b n}$ in Eqs. (1) and (2) yields
$$\begin{array}{l}\mathbf{E}_{A B}=\mathbf{I}_{a n} \mathbf{Z}_{1}-\left[-\left(\mathbf{I}_{a n}+\mathbf{I}_{c n}\right)\right] \mathbf{Z}_{2} \\\mathbf{E}_{B C}=-\left(\mathbf{I}_{a n}+\mathbf{I}_{c n}\right) \mathbf{Z}_{2}-\mathbf{I}_{c n} \mathbf{Z}_{3} \end{array}$$
which are rewritten as
$$\begin{array}{l}\mathbf{E}_{A B}=\mathbf{I}_{a n}\left(\mathbf{Z}_{1}+\mathbf{Z}_{2}\right)+\mathbf{I}_{c n} \mathbf{Z}_{2} \\\mathbf{E}_{B C}=\mathbf{I}_{a n}\left(-\mathbf{Z}_{2}\right)+\mathbf{I}_{c n}\left[-\left(\mathbf{Z}_{2}+\mathbf{Z}_{3}\right)\right]\end{array}$$
Using determinants, we have
\begin{aligned}\mathbf{I}_{a n} &=\frac{\left|\begin{array}{cc}\mathbf{E}_{A B} & \mathbf{Z}_{2} \\\mathbf{E}_{B C} & -\left(\mathbf{Z}_{2}+\mathbf{Z}_{3}\right)\end{array}\right|}{\left|\begin{array}{cc}\mathbf{Z}_{1}+\mathbf{Z}_{2} & \mathbf{Z}_{2} \\-\mathbf{Z}_{2} & -\left(\mathbf{Z}_{2}+\mathbf{Z}_{3}\right)\end{array}\right|} \\&=\frac{-\left(\mathbf{Z}_{2}+\mathbf{Z}_{3}\right) \mathbf{E}_{A B}-\mathbf{E}_{B C} \mathbf{Z}_{2}}{-\mathbf{Z}_{1} \mathbf{Z}_{2}-\mathbf{Z}_{1} \mathbf{Z}_{3}-\mathbf{Z}_{2} \mathbf{Z}_{3}-\mathbf{Z}_{2}^{2}+\mathbf{Z}_{2}^{2}} \\\mathbf{I}_{a n} &=\frac{-\mathbf{Z}_{2}\left(\mathbf{E}_{A B}+\mathbf{E}_{B C}\right)-\mathbf{Z}_{3} \mathbf{E}_{A B}}{-\mathbf{Z}_{1} \mathbf{Z}_{2}-\mathbf{Z}_{1} \mathbf{Z}_{3}-\mathbf{Z}_{2} \mathbf{Z}_{3}}\end{aligned}
Applying Kirchhoff's voltage law to the line voltages:
$$\mathbf{E}_{A B}+\mathbf{E}_{C A}+\mathbf{E}_{B C}=0 \text { or } \mathbf{E}_{A B}+\mathbf{E}_{B C}=-\mathbf{E}_{C A}$$
Substituting for $\left(\mathbf{E}_{A B}+\mathbf{E}_{C B}\right)$ in the above equation for $\mathbf{I}_{a n}$ :
$$\mathbf{I}_{a n}=\frac{-\mathbf{Z}_{2}\left(-\mathbf{E}_{C A}\right)-\mathbf{Z}_{3} \mathbf{E}_{A B}}{-\mathbf{Z}_{1} \mathbf{Z}_{2}-\mathbf{Z}_{1} \mathbf{Z}_{3}-\mathbf{Z}_{2} \mathbf{Z}_{3}}$$
and
$$\bbox[10px,border:1px solid grey]{\mathbf{I}_{a n}=\frac{\mathbf{E}_{A B} \mathbf{Z}_{3}-\mathbf{E}_{C A} \mathbf{Z}_{2}}{\mathbf{Z}_{1} \mathbf{Z}_{2}+\mathbf{Z}_{1} \mathbf{Z}_{3}+\mathbf{Z}_{2} \mathbf{Z}_{3}}} \tag{4}$$
In the same manner, it can be shown that
$$\bbox[10px,border:1px solid grey]{\mathbf{I}_{c n}=\frac{\mathbf{E}_{C A} \mathbf{Z}_{2}-\mathbf{E}_{B C} \mathbf{Z}_{1}}{\mathbf{Z}_{1} \mathbf{Z}_{2}+\mathbf{Z}_{1} \mathbf{Z}_{3}+\mathbf{Z}_{2} \mathbf{Z}_{3}}} \tag{5}$$
Substituting Eq. (5) for $\mathbf{I}_{c n}$ in the right-hand side of Eq. (2), we obtain
$$\bbox[10px,border:1px solid grey]{\mathbf{I}_{b n}=\frac{\mathbf{E}_{B C} \mathbf{Z}_{1}-\mathbf{E}_{A B} \mathbf{Z}_{3}}{\mathbf{Z}_{1} \mathbf{Z}_{2}+\mathbf{Z}_{1} \mathbf{Z}_{3}+\mathbf{Z}_{2} \mathbf{Z}_{3}}}$$

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