Y Connection Style

In a star or wye (Y) connection style, one terminal of each phase of the source or load is connected to the corresponding terminals of the other two phases, making it a neutral point, and leaving three terminals free for connection to the rest of the circuit.

• All of the negative terminals are connected together, all the positive terminals are free

---

Voltage and Current Relationships

Relationship	Value
Phase-to-neutral and line-to-line voltage magnitudes	$V_{LL}=\sqrt{3}{V}_{\text{phase}}$
Phase-to-neutral and line-to-line voltage phase angles	${\vec{V}}_{xy}$ leads ${\vec{V}}_{xn}$ by $30°$, e.g. ${\vec{V}}_{ab}$ leads ${\vec{V}}_{an}$ by $30°$
Phase and line current phasors	${\vec{I}}_L={\vec{I}}_{\text{phase}}$

The derivations for these values are shown below.

Phase-to-Neutral Voltage

The phasor representation of the voltages of the three phases with respect to the neutral point $N$ are as follows:

$$\begin{array}{rl}{\vec{V}}_{AN}& =V_{\text{phase}}\angle 0°\\ {\vec{V}}_{BN}& =V_{\text{phase}}\angle -120°\\ {\vec{V}}_{CN}& =V_{\text{phase}}\angle -240°=V_{\text{phase}}\angle +120°\end{array}$$

where $V_{\text{phase}}$ is the rms value of the phase-to-neutral voltage and phase-A voltage has been taken as a reference.

Line-to-line Voltage

From the phasors of phase-to-neutral voltages, we can find the phasors of line-to-line voltages.

From Fig. 4-6, the phasors of the line-to-line voltages can be found as:

$$\begin{array}{rl}{\vec{V}}_{AB}& =V_{LL}\angle 30°\\ {\vec{V}}_{BC}& =V_{LL}\angle -90°\\ {\vec{V}}_{AB}& =V_{LL}\angle 150°\end{array}$$

where $V_{LL}=\sqrt{3}{V}_{\text{phase}}$.

The above can be derived by doing vector arithmetic. For example, as shown in the diagram above, we have:

$$\begin{array}{rl}{\vec{V}}_{AB}& ={\vec{V}}_{AN}-{\vec{V}}_{BN}\\ & =V-V(\cos (-120°)+j\sin (-120°))\\ & =\frac{3}{2}V+j\frac{\sqrt{3}}{2}V\\ |{\vec{V}}_{AB}|& =\sqrt{{\left(\frac{3}{2}V\right)}^2+{\left(\frac{\sqrt{3}}{2}V\right)}^2}=\sqrt{3}V\\ {\theta }_{AB}& ={\tan }^{-1}\left(\frac{\sqrt{3}}{2}/\frac{3}{2}\right)=30°\end{array}$$

---

Loads and Currents

The relationship between the phase and line voltages will be the same as that for a 3-phase Y-connected source. Let’s look at the currents in the three phases.

We can calculate the currents. For example:

$$\begin{array}{rl}{\vec{I}}_a& =\frac{{\vec{V}}_{an}}{Z_{\text{phase}}}\\ & =\frac{V_{\text{phase}}\angle 0°}{Z_{\text{phase}}\angle \theta }\\ & =\frac{V_{\text{phase}}}{|Z_{\text{phase}}|}\angle -\theta \\ & =I_{\text{phase}}\angle -\theta \end{array}$$

Where $\theta$ is the phase angle of the impedance $Z_{\text{phase}}$.

Then, we also have:

$$\begin{array}{r}{\vec{I}}_b=\frac{{\vec{V}}_{bn}}{Z_{\text{phase}}}={\vec{I}}_{\text{phase}}\angle -120°-\theta \\ {\vec{I}}_c=\frac{{\vec{V}}_{cn}}{Z_{\text{phase}}}={\vec{I}}_{\text{phase}}\angle +120°-\theta \end{array}$$

where $Z_{\text{phase}}$ is the impedance per phase and $I_{\text{phase}}$ is the rms value of the phase current. In the Y-connection style, the current in each line is equal to the current in the corresponding phase. Thus, the line current is equal to the phase current, i.e.

$$I_{\text{line}}=I_{\text{phase}}$$

The same relationships hold true for the currents in a 3-phase balanced Y-connected voltage source.

Phase current vs Line current

• Phase current: The current that flows through each impedance (load) connected between a phase and the neutral point.
• Line current: The current that flows in the conductors connecting the source to the load.

Note that in a 3-phase Y-connected balanced system:

$$\begin{array}{rl}{\vec{V}}_{AN}+{\vec{V}}_{BN}+{\vec{V}}_{CN}=0& \text{or}{\vec{V}}_{an}+{\vec{V}}_{bn}+{\vec{V}}_{cn}=0\\ {\vec{V}}_{AB}+{\vec{V}}_{BC}+{\vec{V}}_{CA}=0& \text{or}{\vec{V}}_{ab}+{\vec{V}}_{bc}+{\vec{V}}_{ca}=0\\ {\vec{I}}_A+{\vec{I}}_B+{\vec{I}}_C=0& \text{or}{\vec{I}}_a+{\vec{I}}_b+{\vec{I}}_c=0\end{array}$$