Balanced loads, in a 3φ system, have identical impedance in each secondary winding (Figure 12). The impedance of each winding in a delta load is shown as Z_{∆} (Figure 12a), and the impedence in a wye load is shown as Z_{y} (Figure 12b). For either the delta or wye connection, the lines A, B, and C supply a 3φ system of voltages.

Figure 12 : 3φ Balanced Loads

In a balanced delta load, the line voltage (V_{L}) is equal to the phase voltage (V_{ø}), and the line current (I_{L}) is equal to the square root of three times the phase current (*√3*I_{ø}).

The below Equation is a mathematical representation of V_{L} in a balanced delta load.

**V _{L }= V_{ø}**

The below Equation is a mathematical representation of I_{L} in a balanced delta load.

**I _{L =}√3I_{ø}**

In a balanced wye load, the line voltage (V_{L}) is equal to the square root of three times phase voltage (*√3V*_{ø}), and line current (I_{L}) is equal to the phase current (I_{ø}).

The below Equation is a mathematical representation of V_{L} in a balanced wye load.

**V _{L }= √3V_{ø}**

The below Equation is a mathematical representation of I_{L} in a balanced wye load.

**I _{L = }I_{ø}**

Because the impedance of each phase of a balanced delta or wye load has equal current, phase power is one third of the total power.

The below Equation is the mathematical representation for phase power (P_{ø}) in a balanced delta or wye load.

**P _{ø }= V_{ø} I_{ø} cosθ**

Total power (P_{T}) is equal to three times the single-phase power.

The below Equation is the mathematical representation for total power in a balanced delta or wye load.

**P _{T }= 3 V_{ø} I_{ø} cosθ**

**In a delta-connected load,**

**In a wye-connected load,**

As you can see, the total power formulas for delta- and wye-connected loads are identical.

Total apparent power (S_{T}) in volt-amperes and total reactive power (Q_{T}) in volt-amperes-reactive are related to total real power (P_{T}) in watts (Figure 13).

Figure 13 : 3φ Power Triangle

A balanced three-phase load has the real, apparent, and reactive powers given by:

**Example 1:**

**Each phase of a delta- connected 3φ AC generator supplies a full load current of 200 A at 440 volts with a 0.6 lagging power factor, as shown in Figure 14.**

Figure 14 : Three-Phase Delta Generator

Find:

- V
_{L} - I
_{L} - P
_{T} - Q
_{T} - S
_{T}

Solution :

**1. Calculate V _{L}**

**V _{L }= V_{ø}**

V_{L }= 440V

**2. Calculate I _{L}**

I_{L =}*√3*I_{ø}

I_{L }= 1.73 x 200

I_{L }= 346 amps

**3. Calculate P _{T}**

P_{T }= *√*3 V_{L} I_{L} cosθ

P_{T }= 1.73 x 440 x 346 x 0.6

P_{T }= 158.2 kW

**4. Calculate Q _{T}**

Q_{T }= *√*3 V_{L} I_{L} sinθ

Q_{T }= 1.73 x 440 x 346 x 0.8

Q_{T }= 210.7 KVR

**5. Calculate S _{T}**

S_{T }= *√*3 V_{L} I_{L}

S_{T }= 263.4 KVA

**Example 2:**

**Each phase of a wye- connected 3φ AC generator supplies a 100 A current at a phase voltage of 240V and a power factor of 0.9 lagging, as shown in Figure 15.**

Figure 15 : Three-Phase Wye Generator

Find:

- V
_{L} - P
_{T} - Q
_{T} - S
_{T}

Solution :

**1. Calculate V _{L}**

V_{L }= *√3V*_{ø}

V_{L }= 1.73 x 240

V_{L }= 415.2 volts

**2. Calculate P _{T}**

P_{T }= *√*3 V_{L} I_{L} cosθ

P_{T }= 1.73 x 415.2 x 100 x 0.9

P_{T }= 64.6 kW

**3. Calculate Q _{T}**

Q_{T }= *√*3 V_{L} I_{L} sinθ

Q_{T }= 1.73 x 415.2 x 100 x 0.436

Q_{T }= 31.3 KVAR

**4. Calculate S _{T}**

S_{T }= *√*3 V_{L} I_{L}

S_{T }= 1.73 x 415.2 x 100

S_{T }= 71.8 KVA