Each coil of three phases has two terminals (one ‘start’ and another ‘finish’) and if individual phase is connected to a separate load circuit, as illustrated in Fig. 7.3, we get a non-interlinked three- phase, six-wire system.
Then, by Ohm’s law, the current in individual phase will be:
IP = EP/ZP … (7.4)
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Where Ip is the phase current, Ep is the phase emf or voltage.
In such a system, each circuit will require two conductors, therefore, 6 conductors in all. Hence such a system will be very complicated and expensive.
In practice non-interlinked 3-phase, 6-wire system is not employed but the phases are interconnected to give 3 or 4-wire, 3-phase system. The two principal methods of connecting the separate phases of three- phase system are the star (or wye) connections and the delta (or mesh) connections.
The interconnections must be carried out in such a way that if closed circuits are formed the sum of instantaneous emfs in them must be zero in order that there shall be no circulating current in these circuits.
Star or Wye (Y) Connected System:
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Star or Wye (Y) connected system is obtained by joining together similar ends, either the start or the finish, the other ends are joined to the line wire, as shown in Fig. 7.4 (a). The common point N at which similar (start or finish) ends are connected is called the neutral or star point.
Ordinarily only three wires are carried to the external circuit giving 3-phase 3-wire star-connected system but sometimes a fourth wire, known as neutral wire is carried to the neutral point of the external load circuit giving 3-phase, 4-wire star-connected system.
The voltage between any line and the neutral point i.e., voltage across the phase winding is called the phase voltage; while the voltage between any two outers is called the line voltage. The neutral point is usually earth-connected.
In Fig. 7.4 (a) positive directions of emfs are taken from star point outwards. The arrowheads on emfs and currents indicate the positive direction and not their directions at any instant. The three phases be either numbered as 1, 2, 3, or a, b, c or R, Y, B. In Fig. 7.4 (a) the three-phases are numbered as usually done; R, Y and B indicate the three natural colours red, yellow and blue respectively.
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It is worthwhile to state here that there are only two possible phase sequences, namely RYB and RBY. By convention sequence RYB is taken as positive and RBY as negative.
In Fig. 7.4 (b) the emfs induced in 3-phases are shown by phasors. In star-connection, there are two windings between each pair of outers and due to joining of similar ends together the emf induced in them are in opposition.
Hence the potential difference between the two outers, known as line voltage, is the phasor difference of phase emfs of the two phases concerned.
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For example, the potential difference between outers R and Y or line voltage ERY is the phasor difference of phase emfs ER and EY or phasor sum of phase emfs FR and (- EY).
Similarly potential difference between outers Y and B
or line voltage EYB = EY – EB = √3 EP in magnitude and potential difference between outers B and R
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Or line voltage, EBR = EB – ER = √3 EP in magnitude
In balanced star (Y) system ERY, EYB and EBR are equal in magnitude and are known as line voltages.
Let each of them be equal to EL
... EL = √3 EP (in balanced star-connected system) … (7.5)
Since in star-connected system each line conductor is connected to separate phase, current flowing through the lines and phases are same:
i.e., Line current, IL = Phase current, IP … (7.6)
Average power, P = 3 EP IP cos ɸ … (7.7)
Substituting EP = EL/√3 and IP = IL for a 3-phase star-connected balanced system in Eq. (7.7) we have:
Total circuit power, P = 3 EL/√3 × IL cos ɸ = √3 EL IL cos ɸ … (7.8)
i.e., Total power, P = √3 × line voltage × line current × power factor.
Total apparent power, S = 3 × apparent power per phase = 3 EP IP = 3 EL/√3 IL = √3 EL IL … (7.9)
Total reactive power, Q = 3 EP IP sin ɸ = 3 EL/√3 = IL sin ɸ = √3 EL IL sin ɸ = P/ cos ɸ sin ɸ = P tan ɸ … (7.10)
It should be noted that in a balanced star- connected system:
(i) Line voltages are 120° apart.
(ii) Line voltages are 30° ahead of the respective phase voltages.
(iii) Line voltages are √3 times of phase voltages.
(iv) Line currents are equal to phase currents.
(v) The angle between line currents and the corresponding line voltages is (30° ± ɸ); +ve for lagging currents and -ve for leading currents.
(vi) True power output = √3 EL IL cos ɸ, where ɸ is the angle between the respective phase current and phase voltage (not between line current and line voltage).
(vii) Apparent power = √3 ELIL
(viii) In balanced system, the potential of neutral or star point is zero because potential at neutral or star point,
ENR + ENY + ENB = 0
Mesh or Delta (Δ) Connected System:
When the starting end of one coil is connected to the finishing end of another coil, as shown in Fig. 7.6 (a), the delta or mesh connection is obtained. The directions of emfs in the coils have been taken as positive from start end to finish end in Fig. 7.6 (a).
From Fig. 7.6 (b) it is obvious that line current is phasor difference of phase currents of two phases concerned. For example the line current in red outer IR will be equal to the phasor difference of phase currents IYR and IRB. The current phasors are shown in Fig. 7.6 (b).
From Figs. 7.6 (a) and 7.6 (b) we have:
Since in delta-connected system, only one phase is included between any pair of line outers, potential difference between the line outers, called the line voltage, is equal to phase voltage.
i.e., Line voltage, EL = Phase voltage, EP … (7.12)
Power output per phase = EP IP cos ɸ where cos ɸ is the power factor of the load
Total power output = 3 EP IP cos ɸ = 3 EL IL/√3 cos ɸ = √3 EL IL cos ɸ … (7.13)
i.e., Total power output = √3 × line voltage × line current × power factor
Apparent power of 3-phase delta-connected system:
= 3 × apparent power per phase
= 3EPIP = 3 EL IL/√3 = √3 ELIL … (7.14)
Total reactive power, Q =3 EP IP sin ɸ = √3 EL IL sin ɸ= P tan ɸ… (7.15)
It should be noted that in balanced delta-connected system:
(i) Line currents are 120° apart.
(ii) Line currents are 30°, behind the respective phase currents.
(iii) Line currents are times phase currents.
(iv) The angle between line currents and corresponding line voltage is (30° ± ɸ) as in star system.
(v) Line voltages are equal to phase voltages.
(vi) Power = √3 EL IL cos ɸ, where ɸ is the phase angle between respective phase current and phase voltage.
(vii) Apparent power = √3 EL IL
(viii) In balanced system the resultant emf in the closed circuit will be zero.
i.e., ERY + EYB + EBR = 0
Hence there will be no circulating current in the mesh it’ no load is connected to the lines.