When the load impedance in the three phases are not equal in magnitude or phase or both, the load is said to be *unbalanced. *If three unequal loads are connected to form a delta and connected across a 3-phase supply the currents in the three loads will not be equal in magnitude and/or phase. The three-phase currents and the line currents will also be unbalanced.

**Example 40.** *A 440 V, 50 Hz, 3-phase supply has delta-connected load having 50 *Ω *between R and **y, *159 *mH between Y and Band *15.9 µF *between B and R. Find;*

*(i) The line current for the sequence RYB. *

*(ii) **The value of star-connected balanced resistors for the same power. *

**Solution.** Phase voltage *(E _{ph}) *= line voltage

*(E*= 440 V

_{L})The potential difference across three-phases are given by

*E*= 440(1 +

_{RY}*j0)*=

*440*

*0°*V

*E _{YB} *= 440( – 0.5 –

*j0.866)*or 440

*–*120° V

*E*= 440(- 0.5 +

_{BR}*j0.866)*or 440 120

*°*V

Impedance* , Z _{RY} *= (50 +

*j0)*= 50 0

*°*Ω

Impedance* , Z _{YB} *= (0 +

*j2rπfL)*= (0 +

*j2π*× 50 × 159 × 10

^{-3})

= (0 +

*j50)*or 50 90

*°*Ω

**Example 41.** A 440 V, 50 Hz, 3-phase supply has delta-connected load having 50 Ω between R and y, 159 mH between Y and Band 15.9 µF between B and R. Find;

*(i) The line current for the sequence RYB.*

*(ii) The value of star-connected balanced resistors for the same power.*

**Solution. **

Phase voltage*(E _{ph})* = line voltage,

*(E*= 400 V

_{L})The potential differences across three-phases are given by:

*E _{RY} *= 400 0°

*E*= 400 – 120°

_{YB}*E*= 400 120°

_{BR}The phase impedances are:

*Z _{RY} * = 31 + j59 = 66.6 62.3°

*Z*= 30 – j40 = 50 – 53.1°

_{YB}*Z*= 80 + j60 = 100 36.9°

_{BR}**Example 42.** *For the unbalanced delta-connected load of Fig 52, find the phase currents, line currents and total power consumed by the load, when phase sequence is:*

*(i) RYB*

*(ii) RBY*

**(AMIE Advanced Electrical M/c Winter, 1998)**

**Solution. **

(i) **Phase sequence RYB : **

The potential differences across three phases are:

*E _{RY} *= 200

*0°*

*E*=

_{YB}*200*

*-120°*

*E*= 200

_{BR}*120°*

*Impedance, Z _{RY} *= 12 +

*j16*= 20

*53.13°*

*Impedance, Z*= 16 +

_{YB}*j12*= 20

*36.8°*

*Impedance, Z*= 8 – j6 =

_{BR}*10*

*– 36.8°*

**Phase sequence RBY:**

Here *E _{RY} * = 200

*E*= 200

_{YB}*E*= 200

_{BR}**Example 43.** *A 3-phase, 400 V, 50 Hz system has the following load connected in delta:*

*Between R and Y lines: a non-reactive resistor of 50 *Ω

*Between Y and B lines: a coil having a resistance of 10 *Ω *and reactance of 30 *Ω

*Between B and R lines: a loss free capacitor of *15 µ *F. *

*Calculate: **(i) **The phase currents **(ii) **The line currents. *

*Assume phase sequence to be RYB. *

**Solution.**

*E _{ph} *=

*E*= 400 V

_{L}The potential difference across the three phases are:

*E _{RY} *= 400

*0°*V

*E*

_{YB}*= 400*

*–*120° V

*E*= 400 120

_{BR}*°*V

**Line Current**

**I _{R}** =I

_{RY}– I

_{BR}= 8 0° – 1.886 210°

= 8 – 1.886 (- 0.866 – *j0.5) *= 8 + 1.633 + *j0.943
*= 9.633 +

*j0.943*=

**9.68**

*5.6°***A. (Ans.)**

**I _{Y}** = I

_{YB}– I

_{RY}= 12.65 – 191.56° – 8 0°

= 12.65(- 0.979 + *j0.2) *– 8 = – 12.38 + *j2.53 *– 8

= – 20.38 + *j2.53 *= **20.53 ****7.07° A. (Ans.)**

I_{B} = I_{BR} – I_{YB} = 1.886 210° – 12.65 – 191.56°

= 1.886 (- 0.866 – j0.5) – 12.65(- 0.979 + j0.2)

= – 1.633 – j 0.943 + 12.38 – j2.53

= 10.747 -j3.473 = **11.29 ****17.4° . **

**(Ans.)**

**Example 45.** *A *400 *V, 3-phase *4 *wire system has the loads *6 – *j8, *6 + j0 *and *4 + j3 Ω *connected
*in

*star. Find:*

*(i)*

*The line currents,*

*(ii)*

*Neutral current, and*

*(iii)*

*Total power.*

**Solution.** The connections are shown in Fig. 56.

= – [23.09 53.13°* *+ 38.48-120°+ 46.18 83.1°

= – [23.09(0.6 + *j0.8) *+ 36.18 (- 0.5 -j0.866) – 46.18 (0.12 + j0.99)]
= – [13.85 + *j*18.47* *– 18.24 – *j*31.59* *+ 5.54 *+.j*45.72]

= – (1.15 + *j*32.6*) *= -1.15. *-j*32.6* *= **32.62 ****-92 ° **

**A. (Ans.)**

Power:

Power taken by phase *R, *

**Example 46**. *A 440/254 volt, *3 *phase, 4-core supplies an unbalanced load represented by the following impedance in ohms connected between R, Y and B phases respectively and the neutral 16 *+ *j12; *14 – *j21 and *25. *The phase sequence is RYE. Calculate the current in each conductor of the cable and the readings on each of the three wattmeters connected in each line and neutral. *

**Solution**. The e.m.fs between the three lines and neutral are given by :

E_{R} = 254(1 + j0)

E_{Y}= 254(- 0.5 – j0.866)

E_{B} = 254(- 0.5 + j0.866) taking ER as the reference vector

Fig.57 (a) Circuit Diagram

Fig.57 (b) Phasor Diagram

**Example 47**. *A *3*-phase, 4-wire system having a 440 V line to line has the following loads connected between the respective lines and neutral: *

Z_{R} = 5 0° Ω, Z_{Y} = 5 37° Ω, Z_{B} = 5 -53° Ω.

*Calculate the current in neutral wire and power taken by each load when phase sequence is: *

*(i) **RYB*

*(ii) **RBY. *

**Solution**. *(i) ***Phase sequence RYB** (Figs. 58 and 59).

The potential differences across the three phases are:

Current in the neutral:

I_{N} = – (I_{R} + I_{Y} + I_{B})

= – (50.8 *0° *+ 50.8 *–*157° + 50.8 173*°*)*
*= – [50.8 + 50.8 (- 0.92 –

*j*0.39)

*+ 50.8(- 0.99 +*

*j*0.122)]

*= – [50.8 – 46.74 –*

*j*19.81

*–*50.3 +

*j*6.2]}

= – (- 46.24 – *j*13.61)* *= 46.24 + *j*13.61* *

**= 48.2 ****16.4 °. (Ans.) **

Now R_{R} = 5 Ω, R_{Y} = 5 cos 37° = 4 Ω,

R_{B} = 5 cos (- 53°) = 3 Ω

Power taken by phase *R, *

P_{R} = 50.8^{2} × 5 = **12903 W. (Ans.)**

P_{Y }= 50.8^{2} × 4 = **10322 W. (Ans.)**

P_{B} = 50.8^{2} × 3 = **7742 W. (Ans.) **

Phase Sequence RBY (Fig. 60):

E_{RN} = 254 0°, E_{YN} = 254 120°

E_{BN} = 254 -120°

**Line currents:**

**Current in the neutral: **

I_{N}=-(I_{R} + I_{Y} + I_{B})

= – (50.8 0° + 50.8 83° + 50.8 – 67°)

= – [50.8 + 50.8(0.122 + j0.992) + 50.8(0.39 – j0.92)] = – (50.8 + 6.197 + j50.4 + 19.81- j46.74)

= – (76.81 + j3.66) = – 76.81 – j3.66

**= 76.89 ****-177.3° A. (Ans.) **

*Power would remain the same because magnitude of branch currents is unaltered. *

**Example 48.** *In a 3-phase, 4-wire system, the line voltage is 400 V and non-inductive loads of*

*5, 4 and 2.5 kW Ware connected between the three lines conductors and the neutral, as shown in Fig. 61.Calculate*

* i. **The current in lines; and** *

*ii. **The current in the neutral conductor.*

**Solution.** Refer Fig. 61.

Phase voltage across each load,

These currents are mutually 120^{o} out of phase as the circuit is purely resistive. Taking I_{R} as the reference vector, line currents may be given as

**I _{R} **= 21.64 0° = 21.64 (1 + j0)

**I**= 17.32 -120° = 17.32 (- 0.5 – j0.866)

_{Y}**I**= 10.82 120° = 10.82 (- 0.5 + j0.866)

_{B}*(ii) ***Current in the neutral wire **(Refer Fig. 62):

**I _{N} **=- (I

_{R }+ I

_{Y }+ I

_{B})

= – [21.64(1- j0) + 17.32 (- 0.5 – j0.866) + 10.82(- 0.5 + j0.866)]

= – [21.64 – 8.66 – j15 – 5.41 + j9.371 = – (7.57 – j5.63)

= -7.57 + j5.63 = **9.434 ****143.4°. (Ans.)**

**Example 49. ***A 3-phase star-connected system with 230 V between each phase and neutral has *

*resistances*

*8 Ω, 10 Ω and 12 Ω respectively in the three phase. Calculate:*

* i. **Current in each phase,*

*ii. **Current in neutral wire, and*

*iii. **Total power absorbed.*

**Solution. **The circuit and vector diagrams are shown in Figs. 63 and 64 respectively.

Potential difference across each phase = 230 V

*(i) ***Phase currents: **

Since the circuit is purely resistive, so currents will be in phase with their respective voltages and will be mutually 120^{0} apart. Taking I_{R}* *as the reference vector (Fig. 64) the line currents are given as:

I_{R} = 28.75 = 28.75(1 + j0)

I_{Y} = 23 -120° = 23 (- 0.5 – j0.866)

I_{B} = 19.17 120° = 19.17 (- 0.5 + j0.866).

*(ii) *

**Current in neutral wire,**

I_{N} = – (I_{R} + I_{Y} + I_{B}) = – [28.75(1 + j0) + 23(- 0.5 – j0.866) + 19.17(- 0.5 + j0.866)

= – [28.75 – 11.5 – j19.92 – 9.58 + j16.6)

= – [7.67 – j3.32) = – 7.67 + j3.32 = **8.357 ****156.6° ****A. (Ans.)**

*(iii) ***Total power absorbed**

= 6612.5 + 5290 + 4409.8 = **16312.3 W. (Ans.)**

**Example 50.** *A 440 V (line-to-line), 3-phase, 4-core supplies an unbalanced load represented by the following impedances in ohms connected between R, Y and B phases respectively and the neutral **8 + j6, 7 – j10.5 and 12.5. Calculate: *

*(i) **Current in each conductor of the cable,*

*(ii) **Current in the neutral wire, and *

*(iii) **Readings of the three wattmeters connected in each line and neutral.*

*The phase sequence is RYB.*

* Solution.*The potential differences across the three phases are:

E_{YN} = 254 -120°

E_{BN} = 254 120°.

**Impedances: **

Z_{R} = 8 + j6 = 10 36.9°

Z_{Y} = 7 – j10.5 = 12.62 – 56.3°

Z_{B} = 12.5 = 12.5 0°.

*(i) ***Phase currents:**

*(ii) ***Current in the neutral wire:**

I_{N} = – (I_{R} + I_{Y} + I_{B})

= – [25.4 -36.9° + 20.12 – 63.7° + 20.32 120°]

= – [25.4(+ 0.8 – j0.6) + 20.12(0.44 – j0.896) + 20.32(- 0.5 + j0.866)]

= – (20.32 – j15.24 + 8.85 – j18.03 – 10.16 + j17.6)

= – (19.01 – j15.67) = – 19.01 + j15.67 = **24.64 ****39.5° A. (An s.)**

*(iii) ***Readings of wattmeters:**

**KIRCHHOFF’S LAWS** (Article 11)

**Example 51.** *Calculate the voltage drop across each of the following three star-connected impedances when they are connected to a 400 V, 3-phase system: *

Z_{R} = j40Ω, Z_{Y} = j2Ω, Z_{B} = – j2Ω.

*Also find the potential of the star-point above ground.*

*The phase sequence is RYB. *

**Solution.** Refer Fig. 67.

Considering E_{RY} as a reference vector. The vector expressions for the three line voltages are:

E_{RY} = 400(1 + j0) V = (400 + j0) V

E_{YB} = 400(- 0.5 – j0.866) = (- 200 – j346.4) V

E_{BR} = 400(- 0.5 + j0.866) = (- 200 + j346.4) V

**Impedances: **

Z_{R} = (0 + j40),

Z_{Y}= (0 + j2)

Z_{B} = (0 – j2)

Now for finding I_{R}, I_{Y} and I_{B}, using the relations:

Voltage drops across impedances:

Voltage drop, E_{R} = I_{R}Z_{R} = (173.2 – j100) × j40 = **(4000 + j6928) V. (Ans.) **

Voltage drop, E_{Y} = I_{Y}Z_{Y} = (3464 – j1800)(j2) = **(3600 + j6928) V. (Ans.) **

Voltage drop, E_{B} = I_{B}Z_{B} = (- 3637.2 + j1900) (- j2) = **(3800 + j7274.4) V. (Ans.) **

Potential of star point above ground:

**Example 52**. *An unbalanced star-connected load shown in Fig. *68, *Z _{R} *==

*j40*Ω,

*Z*= –

_{Y}*j40*Ω,

*Z*= 40 Ω

_{B}*is connected across a balanced 400 V, 3-phase supply. Find the reading of the wattmeter connected in Y-phase.*

*The phase sequence is R *→ *Y *→ *B. *

**Solution.** Refer Fig. 68.

Taking *E _{RY} *as reference vector

E_{Ry} = 400 0° = 400 (1 + j0)

E_{YB} = 400 -120^{0} == 400 (- 0.5 – j0.866)

E_{BR} == 400 120° == 400(- 0.5 + j0.866)

Impedances are

Z_{R} == (0 + j40), Z_{Y} = (0 – j40), Z_{B} = (40 + j0)

Voltage drop, E_{R }= I_{R}Z_{R} = (1.34 – j5) × j40 = (200 + j53.6) V

Voltage drop, E_{Y} = I_{Y}Z_{Y} = (- 1.34 – j5) × (- j40) = (- 200 + j53.6) V

Voltage drop, E_{B} = I_{B}Z_{B} = j10 × 40 = (0 + j400) V

Voltage cross potential coil of wattmeter

= voltage between supply neutral and load neutral

Current through wattmeter current coil x conjugate of voltage across wattmeter potential coil

= I_{Y} × (- j169) = (- 1.34 – j5) × (- j169) = (- 845 + j226.5)

**STAR-DELTA CONVERSION**

**Example 53.** *An unbalanced star-connected load has branch impedances of Z _{R} = 10 *

*30° Ω, Z*

_{Y}= 20*45° Ω and Z*

_{B}= 20*60° Ω and is connected across a balanced 3-phase, 3-wire supply of 440 V. Find:*

*i. **Line currents, and*

*ii.**Voltage across each impedance*

*Use star=delta conversion method.*

**Solution. **Taking E_{RY} are reference vector

Then E_{RY} = 440 0°

E_{YB} = 440 -120°

E_{BR} = 440 120°

Refer Fig.69

Fig. 69 (a)

Fig. 69 (b)

Also Z_{R} = 10 30° (given)

Z_{Y}= 20 45° (given)

Z_{B} = 20 60° (given)

**Impedances of equivalent delta-connected load **(Refer Fig.69):

*(i) *

**Line currents:**

**I _{R}** = I

_{RY}– I

_{BR}= (9.34 – j6.3) – (3.625 + j10.66)

= 5.715 – j16.96 =

**17.89**

**-71.4°**

**A. (Ans.)**

**I**= I

_{Y}_{YB}– I

_{RY}= (- 5.62 + j0.37) – (9.34 – j6.3)

= -14.96 + j6.67 = **16.4 ****156° ****A. (Ans.).**

I_{B} = I_{BR} – I_{YB} = (3.625 + j10.66) – (- 5.62 + j0.37) .

= 9.245 + j10.29 = **13.833 ****48° ****A. (Ans.)**

These are the currents in the phases of the star-connected unbalanced load. Let us find voltage

drop across each star-connected branch impedance.

*(ii ) *

**Voltage across impedances :**

**Voltage across** Z_{R} = I_{R} × Z_{R}

= 17.89 -71.4° × 10 30° = **178.9 ****– ****41.4° V. (Ans.)**

**Voltage across** Z_{Y} = I_{Y} × Z_{Y} = 16.4 156° × 20 45°

**= 328 L210° V. (Ans.) **

**Voltage across** Z_{B} = I_{B} × Z_{B} = 13.833 48° × 20 60°

**= 276.67 ****108° V. (Ans.) **

The phasor diagram is shown in Fig. 70.

**Example 54**. *The branch impedances of an unbalanced star-connected load are: Z _{R} *=

*20*

*30° Ω, Z*=

_{Y}*20*

*–*45° Ω,

*Z*=

_{B}*40*

*60°*Ω.

*It is connected across a balanced 3-phase, 3-wire supply of 200 V. Find:*

*(i) **Line currents, and*

*(ii) **Voltage across each impedance. *

*(iii) **Use star-delta conversion method. *

**Solution.** The unbalanced star-connected load and its equivalent delta-connected load are shown in Fig. 71.

Fig. 71 (a)

Fig. 71 (b)

Z_{R} = 20 30° Ω, Z_{Y} = 20 – 45° Ω, Z_{B} = 40 60° Ω

Now Z_{R}Z_{Y} + Z_{Y}Z_{B} + Z_{B}Z_{R}

= (20 300)(20 – 45°) + (20 – 45°)(40 60°) + (40 60°)(20 30°)

= 400 – 15° + 800 15° + 800 90°

= 400(0.966 – j0.26) + 800(0.966 + j0.24) + j800

= 386.4 – j104 + 772.8 + j208 + j800 = 1159.2 + j904 = 1470 38°

Equivalent impedances of delta-connected load are:

*(i) ***Line Currents:**

I_{R} = I_{RY }– I_{BR}

= (5.04 + j2.04) – (2.17 + j1.63) = 2.87 + j0.41 = 2.9 8.1 ° A. (Ans.)

I_{Y} = I_{YB} – I_{RY} = (- 1.67 – j2.15) – (5.04 + j2.04)

= – 6.71 – j4.19 = 7.91 -148° A. (Ans.)

I_{B} = I_{BR} – I_{YB} = (2.17 + j1.63) – (-1.67 – j2.15)

= 3.84 + j3.78 = 5.39 44.5° A. (Ans.)

These are the currents in the phases of star-connected unbalanced load. Let us find the drop across each star-connected branch impedance.

* **(ii) ***Voltage drop across impedances:**

Voltage drop across Z_{R} = I_{R}Z_{R} = 2.9 8.1° × 20 30° = 58 38.1° V. (Ans.)

Voltage drop across Z_{Y} = I_{Y}Z_{Y} = 7.91 -148° × 20 – 45° = 158.2 -193° V. (Ans.)

Voltage across Z_{B} = I_{B}Z_{B} = 5.39 44.5° × 40 60° = 215.6 104.5° V. (Ans.)

**Example 55**. *A star-connected load consisting of non-inductive resistors of 30, **12 and **20 ohms connected to the R, **Y and B lines respectively is fed by a 300 V (line), 3-phase supply. The phase sequence is RYB. Calculate the voltage across each resistor.*

Fig.72

**Using star-delta conversion method,**

**Impedances are :**

Z_{R} = 30 Ω, Z_{Y} = 12 Ω , Z_{B} = 20 Ω

Now Z_{R}Z_{Y} + Z_{Y}Z_{B} + Z_{BR} = 30 × 12 + 12 × 20 + 20 × 30 = 360 + 240 + 600 = 1200 Ω

Equivalent impedances of delta-connected load are:

Each current is in phase with its own voltage because the load is purely resistive.

To find line currents for delta-connection using vector algebra and choosing *E _{RY} *as the reference axis, we get

I_{RY }= 5 + j0

I_{YB} = 7.5(- 0.5 – j0.866) = – 3.75 – j6.5

I_{BR} = 3(- 0.5 + j0.866) = – 1.5 + j2.6

**Line currents for delta-connection are: **

I_{R} = I_{RY} – I_{BR}

= (5 + j0) – (- 1.5 + j2.6) = (6.5 – j2.6) or 7 A in magnitude

I_{Y} = I_{YB} -I_{RY}

= (- 3.75 – j6.5) – (5 + j0) = – 8:75 – j6.5 or 10.9 A in magnitude

I_{B} = I_{BR} =I_{YB} = (- 1.5 + j2.6) – (- 3.75 – j6.5)

= 2.25 – j9.1 or 9.37 A in magnitude

These line currents for delta-connection are the phase currents for star-connection.

Voltage drop across each limb of star-connected load is :

**Voltage drop across Z _{R}** = I

_{R}Z

_{R}= (6.5 – j2.6) × 30 = 195 – j78 or

**210 V.**

**(Ans.)**

**Voltage drop across**

**Z**= I

_{Y}_{Y}Z

_{Y}= (- 8.75 -j6.5) × 12 = – 105 – j78 or

**130.8 V.**

**(Ans.)**

**Voltage drop across Z**= I

_{B}_{B}Z

_{B}= (2.25 – j9.1) × 20 = 4.5 – j182 or

**182.05 V.**

**(Ans.)**

**Example 56.** *A 400 V, 3-phase, 3-wire symmetrical system R _{YB} supplies a star-connected load whose branch circuit impedances are: *

*Determine the current in each branch.
Take the phase sequence as RYB.
*

**Solution**. The circuit is shown in Fig. 73.

Let us solve this question by *four *possible ways of tackling 3-wire unbalanced star-connected load problems.

Fig.73

**1. ****By using Kirchhoff’s law :**

We know that

Now Z_{R}Z_{Y} + Z_{Y}Z_{B} + Z_{B}Z_{R}

= (10 30°) (20° 60°) + (20 60°) (5 -90°) + (5 – 90°) (10 30°)

= 200 90° + 100 -30° + 50 -60°

= j200 + (86.6 – j50) + (25 – j43.3)

= 111.6 +j106.7 = 154.4 43.7°

E_{RY}Z_{B} – E_{BR}Z_{Y} = 400 × 5 -90° – 400 120° × 20 60°

= 2000 -90° – 8000 180° = – j2000 – (- 8000)

= 8000 – j2000 = 8246 -14°

E_{YB}Z_{R} – E_{RY}Z_{B} = 400 -120° × 10 30° – 400 0° × 5-90°

= 4000 -90° – 2000 -90°

=-j4000 + j2000 = – j2000 = 2000 -90°

E_{BR}Z_{Y} – E_{YB}Z_{R} = 400 120^{o} × 20 60° – 400 -120° × 10 30°

=8000180^{0} – 4000 -90° = – 8000 + j4000 = 8944 153.4°

Current in each branch [from (i), (ii) and (iii)]

** 2. By using Star-delta conversion method: **The unbalanced star-connected load and its equivalent delta-connected load are shown in Fig 74.

Fig.74 (a)

Fig.74 (b)

Now, Z_{R}Z_{Y} + Z_{Y}Z_{B} +Z_{B}Z_{R} = 154.4 43.7° [Already calculated]

Equivalent delta-connected impedances are:

**Line currents for delta-connection are: **

I_{R} =I_{RY} – I_{BR}

= (- 8.95 – j9.36) – (- 37.15 + j35.78)

= 28.5 – j15.11 = **53.38 ****57.7° A. (Ans.)**

I_{Y} =I_{YB} – I_{RY}

= (- 17.89 – j18.72) – (- 8.95 + j9.36)

= -8.94 – j9.36 = 12.94 133.7°

I_{B}=I_{BR }– I_{YB}

= (- 37.45 +j35.78) – (- 17.89 – j18.72) = – 19.56 + j54.5 = 57.9 109.7°

ƩI = I_{R} + I_{Y} + I_{B} = 0 ……… as a check.

**3. By using Maxwell’s Loop Current Method: **

Fig. 75 shows the loop or mesh currents.

Fig. 75

It may be noted that I_{R} = I_{1}

I_{Y }= I_{2 }– I_{1}

and I_{B }= -I_{2}

Considering drops across R and Y arms, we get

I_{1}Z_{R} + Z_{Y}(I_{1} – I_{2}) = E_{RY}

or I_{1}(Z_{R} + Z_{Y}) – I_{2}Z_{Y} = E_{RY} …*(i)*

Similarly, considering the arms Y and B, we have

(I_{2} – I_{1}) Z_{Y} + I_{2}Z_{B} = E_{YB}

-I_{1}Z_{Y} + I_{2}(Z_{Y} + Z_{B}) = E_{YB}

I_{R} = I_{1} = **53.5 ****– 57.7° ****A. (Ans.)**

I_{Y} = I_{2} – I_{1} = (19.58 – j54.7) – (28.59 – j45.22)

= – 90 – j9.48 = **13.07 ****-133.5° ****A. (Ans.)**

I_{B} = -I_{2} = – 58.08 -70.3° = **58.08 ****109.7° ****A. (Ans.)**

**4. Using Milliman’s Theorem:**

Refer Figs. 76 and 77.

According to Milliman’s theorem, the voltage of the *load *star point 0′ with respect to the star point or neutral 0 of the generator or supply (normally zero potential) is given by

where *E _{RO}, E_{YO} *and

*E*are the phase voltages of the generator or 3-phase supply.

_{BO}Voltage across each phase of the load is

E_{RO´} = E_{RO} – E_{O´O}

E_{YO´} = E_{YO} – E_{O´O}

E_{BO´} = E_{BO} -E_{O´O}

Now, I_{RO´} = (E_{RO} -E_{O´O})

I_{YO´} = (E_{YO} – E_{O´O}) Y_{Y}

I_{BO´} = (E_{BO} – E_{O´O}) Y_{B}

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