(i) The convention adopted to identify each of the phase voltage is: R—Red, Y—Yellow and B—blue, as shown in Figure 18.2.

(ii) The phase-sequence is given by the sequence in which the conductors pass the point initially taken by the red conductor. The national standard phase sequence is R, Y, B.

(i) A star-connected load is shown in Figure 18.3 where the three line conductors are each connected to a load and the outlets from the loads are joined together at N to form what is termed the neutral point or the star point.

(ii) The voltages, V_R, V_Y and V_B are called phase voltages or line to neutral voltages. Phase voltages are generally denoted by V_P.

(iii) The voltages, V_RY, V_YB and V_BR are called line voltages.

(iv) From Figure 18.3 it can be seen that the phase currents (generally denoted by I_p) are equal to their respective line currents I_R, I_Y and I_B, i.e. for a star connection:

$I_L=I_P$

(v) For a balanced system: I_R = I_Y = I_B, V_R = V_Y = Y_B, V_RY = V_YB = V_BR, Z_R = Z_Y = Z_B and the current in the neutral conductor, I_N = 0. When a star connected system is balanced, then the neutral conductor is unnecessary and is often omitted.

(vi) The line voltage, V_RY, shown in Figure 18.4(a) is given by V_RY = V_R-V_Y. (V_Y is negative since it is in the opposite direction to V_RY.) In the phasor diagram of Figure 18.4(b), phasor V_Y is reversed (shown by the broken line) and then added phasorially to V_R, (i.e. V_RY = V_R + (-V_Y)). By trigonometry, or by measurement, V_RY = √3V_R, i.e. for a balanced star connection:

$V_L=\sqrt{3V_P}$

    A phasor diagram for a balanced, three-wire, star-connected, 3-phase load having a phase voltage of 240 V, a line current of 5 A and a lagging power factor of 0.966 is shown in Figure 18.5 The phasor diagram is constructed as follows:

(i) Draw V_R = V_Y = V_B = 240 v and spaced 120° apart.

(ii) Power factor = cos ϕ = 0.966 lagging. Hence the load phase angle is given by arccos 0.966, i.e. 15° lagging. Hence

$I_R=I_Y=I_B=5A$

    lagging V_R, V_Y and V_B respectively by 15°.

(iii) V_RY = V_R−V_Y (phasorically). By measurement, V_RY = 415 V (i.e. √3(240)) and leads V_R by 30°. Similarly, V_YB = V_Y−V_B and V_BR = V_B−V_R.

(vii) The star connection of the three phases of a supply, together with a neutral conductor, allows the use of two voltages — the phase voltage and the line voltage. A 4-wire system is also used when the load is not balanced. The standard electricity supply to consumers in Great Britain is 415/240 V, 50 Hz, 3-phase, 4-wire alternating voltage, and a diagram of connections is shown in Figure 18.6

(i) A delta (or mesh) connected load is shown in Figure 18.7 where the end of one load is connected to the start of the next load.

(ii) From Figure 18.7, it can be seen that the line voltages V_RY, V_YB and V_BR are the respective phase voltages, i.e. for a delta connection:

$V_L=V_P$

(iii) Using Kirchhoff’s current law in Figure 18.7,

$I_R=I_{RY}-I_{BR}=I_{RY}+\left(-I_{BR}\right).$

    From the phasor diagram shown in Figure 18.8, by trigonometry or by measurement, I_R = √3 I_RY, i.e for a delta connection:

$I_L=\sqrt{3I_P}$

(i) One-wattmeter method for a balanced load

    Wattmeter connections for both star and delta are shown in Figure 18.9 Total power = 3 X wattmeter reading.

(ii) Two-wattmeter method for balanced or unbalanced loads A connection diagram for this method is shown in Figure 18.10 for a star-connected load. Similar connections are made for a delta-connected load.

    Total power = sum of wattmeter readings = P₁ + P₂.

    The power factor may be determined from:

$\tan \varphi =\sqrt{3}\frac{\left(P_1-P_2\right)}{\left(P_1+P_2\right)}$

    It is possible, depending on the load power factor, for one wattmeter to have to be ‘reversed’ to obtain a reading. In this case it is taken as a negative reading.

(iii) Three-wattmeter method for a three-phase, 4-wire system for balanced and unbalanced loads (see Figure 18.11)

$\text{Total power=}{\text{P}}_1+{\text{P}}_2+{\text{P}}_3$

(i) Loads connected in delta dissipate three times more power than when connected in star to the same supply.

(ii) For the same power, the phase currents must be the same for both delta and star connections (since power = 3I²_pR_p), hence the line current in the delta-connected system is greater than the line current in the corresponding star-connected system. To achieve the same phase current in a star-connected system as in a delta-connected system, the line voltage in the star system is √ times the line voltage in the delta system.

    Thus for a given power transfer, a delta system is associated with larger line currents (and thus larger conductor cross-sectional area) and a star system is associated with a larger line voltage (and thus greater insulation).

(i) For a given amount of power transmitted through a system, the three-phase system requires conductors with a smaller cross-sectional area. This means a saving of copper (or aluminium) and thus the original installation costs are less.

(ii) Two voltages are available (see para. 7).

(iii) Three-phase motors are very robust, relatively cheap, generally smaller, have self-starting properties, provide a steadier output and require little maintenance compared with single-phase motors.