### Three Phase Systems

**At the end of this week, the students are expected to:**

- Explain how 3-phase e.m.fs are produced
- Distinguish between star and delta three-phase system
- Derive the relationship between line and phase values of voltages and currents in a star and delta connected windings

**2.5 GENERATION OF THREE PHASE E.M.F’s**

In figure 2.4, three similar coils (A,B, and C) are displaced from one another by 120 electrical degrees. If the coils are rotated within the magnetic field, emf would be induced or generated in the three coils. It is evident that counterclockwise rotation results in coil sides A, B, and C in the order A-B-C. The result for the three coils is as shown in fig 2.3. Voltage B is 120 electrical degrees later than A, and C is 2400 later. Changing the direction of rotation would result in A-C-B, which is called the ACB phase sequence.

**2.6 DIFFERENCE BETWEEN STAR AND DELTA 3–PHASE**

**SYSTEM**

**2.6.1 Star-Connected 3-Phase System**

- Neutral wire is available
- Phase current = line current
- Phase voltage = line voltage √3
- It can handle both lighting and power loads

### 2.6.2 Delta-Connected 3-Phase system

- Neutral wire is not present
- Phase current = line current
- Phase voltage = line voltag
- It can handle power loads only

### 2.7 DERIVATION OF THE RELATIONSHIP BETWEEN LINE AND PHASE VALUES OF VOLTAGES AND CURRENT IN A STAR AND DELTA CONNECTED WINDINGS

**2.7.1 Line voltages and phase voltages in a star connected windings **

Fig 2.5: (a) Star connection of a 3-Φ circuit

### Fig 2.6: phasor diagram of a star connected load

Consider the star connection of a three phase circuit shown in fig 2.5. It phasor diagram is as shown in fig 2.6. To obtain the line voltages we proceed as under: Let VRY = line voltage between red phase and yellow phase

VYB = line voltage between yellow phase and blue phase

VBR = line voltage between blue phase and red phase

VR = voltage across the red phase

VY = voltage across the yellow phase

VB = voltage across the blue phase

Thus, VYB, VBR and VYB are called line voltages, while VB, VR and VY are called phase voltages.

The p.d between line 1 and 2 in (fig 2.5) is VRY. Hence, VRY is found by

compounding VR and VY reversed and its value is given by the diagonal of the

0parallelogram of fig 2.6. Obviously, the angle between VR and VY reversed is 60.

**The parallelogram is shown in fig 2.7 below. **

**Fig 2.7: parallelogram of fig 2.6 **

From fig 2.7, ox = ½ VRY (2.1)

Also, ox = VR cos 300 (2.2)

Equating (2.2) to equation (2.1) gives

VRY = 2 (VRcos300)

= √3 VR = √3 Vph

Considering that the system is balanced,

– VYB = √3 Vph

also, VBR = √3 Vph

Now VRY = VYB = VBR = line voltage, say VL. Hence, in star connection

VL = √3Vph (2.3)

### 2.7.2 Line currents and phase currents in a star connected windings

**Fig 2.8: star connection of three-phase circuit **

### Consider fig 2.8.

Let IR = current flowing through the red phase

IY = current flowing through the yellow phase and

IB = current flowing through the blue phase

IR, IY and IB are called phase current (Iph). It is seen that each of the phase currents is equal to the current flowing through the respective lines. Thus, the current flowing through the respective lines is known as the line current (IL)

Hence, in star connection, line current = phase current.

– IL = Iph (2.4) 2.7.3 Line currents and phase currents in a delta connected windings

**Fig 2.9 Delta connection of a three phase circuit **

Fig 2.10 Phasor diagram of a delta connection of a three phase circuit

Consider the delta connection of a three phase circuit shown in fig 2.9. It phasor diagram is as shown in fig 2.10. To obtain the line currents we proceed as under: Let I1 = IR – IB

I2 = IY – IR

I3 = IB – IY

I1, I2 and I3 are called line currents

Line current I1 is also found by compounding IR and IB reversed as shown in fig 2.10. Its value is given by diagonal of fig 2.10. The parallelogram of fig 2.10 is as shown in fig 2.11.

Fig 2.11: parallelogram of fig 2.10 |

From fig 2.11,

Ox = ½ I1 (2.5)

Also, ox = IB cos300 (2.6)

From equation (2.6) and (2.6), ½ I1 = √3/2 IB If IR = IY = IB = phase current (Iph), then

Current in line no.1 is I1 = √3Iph Current in line no.2 is I2 = √3Iph and

Current in line no.3 is I3 = √3 Iph Since all the line currents are equal in magnitude, i.e I1 = I2 = I3 = IL – IL = √3Iph (2.7) **2.7.4 **Line voltages and phase voltages in a delta connected windings It is seen from fig 2.9 that there is only one phase winding completely included between any pair of terminals. Hence, in delta connection, the voltage between any pair of lines is equal to the phase voltage of the phase winding connected between the two lines considered. Hence, for a balanced system, VRY = VYB = VBR = line voltage VL. Then, it is seen that

VL = Vph (2.8)

## **THANKS FOR VISIT TO PROJECTandMATERIALS **

### *Not just projectandmaterials but the no.1 ultimate source to all your educational info resources and project materials of various discipline.*

*Not just projectandmaterials but the no.1 ultimate source to all your educational info resources and project materials of various discipline.*

## RESEARCH COURSES

### **All research courses listed on our website is for educational and Research purpose, Locate your course title and access the research materials. **

### * The research courses comprises of two categories. ***The general course materials.**

*Note:* success occur when opportunity meets preparation…go for success, go for the feature, study to show your self approve.

*The research courses comprises of two categories.*

**The general course materials.***Note:*success occur when opportunity meets preparation…go for success, go for the feature, study to show your self approve.