12.6 Balanced Delta-Wye Connection

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12.6 Balanced Delta-Wye Connection

A balanced ∆-Y system consists of a balanced ∆-connected source feeding a balanced Y-connected load.

Consider the ∆-Y circuit in Fig. 12.18. Again, assuming the abc sequence, the phase voltages of a delta-connected source are

$$\mathbf{V}_{ab} = V_p \underline{\hat{\mathbf{O}}^{\circ}}, \qquad \mathbf{V}_{bc} = V_p \underline{\hat{\mathbf{O}}^{\circ}} = V_p \underline{\hat{\mathbf{O}}^{\circ}}$$

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$$\mathbf{V}_{ca} = V_p \underline{\hat{\mathbf{O}}^{\circ}} \tag{12.34}$$

These are also the line voltages as well as the phase voltages.

We can obtain the line currents in man y ways. One way is to apply KVL to loop aANBba in Fig. 12.18, writing

$$-\mathbf{V}_{ab} + \mathbf{Z}_{Y}\mathbf{I}_{a} - \mathbf{Z}_{Y}\mathbf{I}_{b} = 0$$

or

Thus,

$$\mathbf{Z}_{Y}(\mathbf{I}_{a}-\mathbf{I}_{b})=\mathbf{V}_{ab}=V_{p}\underline{\int_{0}^{\circ}}$$

^{Ia −} **^I**b = Vp⧸ 0° _____ ZY (12.35)

Figure 12.18 A balanced ∆-Y connection.

But Ib lags Ia by 120°, since we assumed the abc sequence; that is, Ib = Ia^⧸−120°. Hence,

$$\mathbf{I}_a - \mathbf{I}_b = \mathbf{I}_a (1 - 1/ -120^\circ)$$

= $\mathbf{I}_a \left( 1 + \frac{1}{2} + j\frac{\sqrt{3}}{2} \right) = \mathbf{I}_a \sqrt{3}/30^\circ$ (12.36)

Substituting Eq. (12.36) into Eq. (12.35) gives

$$I_a = \frac{\left(V_p / \sqrt{3}\right) / -30^{\circ}}{Z_Y}$$

(12.37)

From this, we obtain the other line currents Ib and Ic using the positive phase sequence, i.e., Ib = Ia^⧸−120°, Ic = Ia^⧸+120°. The phase currents are equal to the line currents.

Another w ay to obtain the line currents is to replace the deltaconnected source with its equi valent wye-connected source, as sho wn

Transforming a ∆-connected source to an equivalent Y-connected source.

per phase, according to Eq. (9.69).

Once the source is transformed to wye, the circuit becomes a wyewye system. Therefore, we can use the equi valent single-phase circuit shown in Fig. 12.20, from which the line current for phase a is

$$I_a = \frac{V_p / \sqrt{3} / -30^{\circ}}{Z_Y}$$

(12.39)

ZY Ia V^p ‒30° √3 + ‒

which is the same as Eq. (12.37).

Alternatively, we may transform the wye-connected load to an equivalent delta-connected load. This results in a delta-delta system, which can be analyzed as in Section 12.5. Note that

$$\mathbf{V}_{AN} = \mathbf{I}_a \mathbf{Z}_Y = \frac{V_p}{\sqrt{3}} \frac{1 - 30^\circ}{\sqrt{3}} \tag{12.40}$$

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$$\mathbf{V}_{BN} = \mathbf{V}_{AN} \frac{1 - 120^\circ}{\sqrt{3}} \qquad \mathbf{V}_{CN} = \mathbf{V}_{AN} \frac{1 + 120^\circ}{\sqrt{3}} \tag{12.40}$$

As stated earlier , the delta-connected load is more desirable than the wye-connected load. It is easier to alter the loads in any one phase of the delta-connected loads, as the individual loads are connected directly across the lines. Ho wever, the delta-connected source is hardly used in practice because any slight imbalance in the phase voltages will result in unwanted circulating currents.

Table 12.1 presents a summary of the formulas for phase currents and voltages and line currents and voltages for the four connections. Students are advised not to memorize the formulas but to understand ho w

TABLE 12.1

	Summary of phase and line voltages/currents for
balanced three-phase systems.	1

Connection	Phase voltages/currents	Line voltages/currents
Y-Y	Van = Vp ⧸ 0°	__ √ Vab = 3 Vp ⧸ 30°
	Vbn = Vp −120° ⧸	Vbc = Vab −120° ⧸
	Vcn = Vp +120° ⧸	Vca = Vab +120° ⧸
	Same as line currents	I Van Z a = ∕ Y
		I I −120° b = ⧸ a
		I I +120° c = ⧸ a
Y- ∆	Van = Vp ⧸ 0°	__ √ Vab = VAB = 3 Vp 30° ⧸
	Vbn = Vp −120° ⧸	Vbc = VBC = Vab −120° ⧸
	Vcn = Vp +120° ⧸	Vca = VCA = Vab +120° ⧸ __
	IAB = VAB Z ∕ ∆	√ I IAB 3 −30° a = ⧸
	IBC = VBC Z ∕ ∆	I I −120° b = ⧸ a
	ICA = VCA Z ∕ ∆	I I +120° c = ⧸ a
- ∆ ∆	Vab = Vp ⧸ 0°	Same as phase voltages
	Vbc = Vp −120° ⧸
	Vca = Vp +120° ⧸	__
	IAB = Vab Z ∕ ∆	√ I a = IAB 3 −30° ⧸
	IBC = Vbc Z ∕ ∆	I b = I −120° ⧸ a
	ICA = Vca Z ∕ ∆	I c = I +120° ⧸ a
∆-Y	Vab = Vp ⧸ 0°	Same as phase voltages
	Vbc = Vp −120° ⧸
	Vca = Vp +120° ⧸	−30° ________ Vp ⧸
	Same as line currents	I a = __ √ 3 Z Y
		I I −120° b = ⧸ a
		I I +120° c = ⧸ a

1 Positive or abc sequence is assumed. they are derived. The formulas can always be obtained by directly applying KCL and KVL to the appropriate three-phase circuits.

A balanced Y-connected load with a phase impedance of 40 + j25 Ω is supplied by a balanced, positive sequence ∆-connected source with a line voltage of 210 V. Calculate the phase currents. Use Vab as a reference.

Solution:

The load impedance is

$$\mathbf{Z}_Y = 40 + j25 = 47.17 / 32^{\circ} \,\Omega$$

and the source voltage is

$$\mathbf{V}_{ab} = 210 \underline{\big/ 0^{\circ}} \, \mathrm{V}$$

When the ∆-connected source is transformed to a Y-connected source,

$$\mathbf{V}_{an} = \frac{\mathbf{V}_{ab}}{\sqrt{3}} \underline{/-30^{\circ}} = 121.2 \underline{/-30^{\circ}} \text{ V}$$

The line currents are

$$\mathbf{I}_a = \frac{\mathbf{V}_{an}}{\mathbf{Z}_Y} = \frac{121.2 \div 30^\circ}{47.12 \div 32^\circ} = 2.57 \div 62^\circ \text{ A}$$

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$$\mathbf{I}_b = \mathbf{I}_a \div 120^\circ = 2.57 \div 178^\circ \text{ A}$$

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$$\mathbf{I}_c = \mathbf{I}_a \div 120^\circ = 2.57 \div 58^\circ \text{ A}$$

which are the same as the phase currents.

In a balanced ∆-Y circuit, Vab = 440⧸ 15° and ZY = (12 + j15) Ω. Practice Problem 12.5 Calculate the line currents.

Answer: 13.224⧸−66.34° A, 13.224⧸+173.66° A, 13.224⧸ 53.66° A.