*Module 2. **A.C. series and
parallel circuits*

**Lesson
9**

**STAR DELTA TRANSFORMATION**

**9.1 Transformations**

Alternating voltage and current can be shown using phasor diagram. The problems can be analysed by using following two mathematical forms:

**9.1.1
Rectangular form **

Consider a voltage phasor V. The magnitude of the phasor is V and having θ angle from reference line OX.

**Fig. 9.1 Voltage phasor
denoted by OV**

** **

Voltage phasor V* *= a
+ jb

Magnitude of phasor

**9.1.2
Polar form**

In polar form the phasor can be represented by the magnitude and the angle with the reference axis.

V = V∠ θ°

**9.2 Conversion Methods **

1. To convert Rectangular to polar form:

Rectangular form V = a + jb

Polar form = V ∠ θ°

2. To convert Polar to Rectangular form:

Rectangular form = V ∠ θ°

a = V cos θ

b = V sin θ

Polar form = a + jb

**9.3
Numerical**

**Q.1
Convert 100 + j 50 to polar form:**

Magnitude =

= 111.80

Phase angle =

= 26.56°

Polar form = 111.80∠ 26.56°

**Q.2
Convert 100****∠**** 14.47°
to rectangular form.**

a = V cosθ

= 100 cos14.47°

= 96.82

b = V sin θ

=100 sin 14.47°

= 25

Rectangular form = 96.82 + j25

**Q.3
Convert 200−j50 and −25−j20 into polar forms.**

1. 200-j50 into polar

Magnitude =

= 206.15

Phase angle

= −14.03°

Polar form = 206.15∠−14.03°

2. −25−j20 into polar

Magnitude =

= 32.01

Phase angle

= 38.65°

Polar form = 32.01∠38.65°

**9.4 ** **Adding
and Subtraction of Phasors**

The rectangular form is the simplest method for addition or subtraction of Phasors. If the phasors are represented in polar form, they should be first converted to rectangular form and then addition or subtraction be carried out.

(i) Addition: For the addition of phasors in the rectangular form, the real components are added together and the complex numbers (j components) are added together. Consider two voltage phasors:

V_{1} = a_{1}
+ jb_{1}; V_{2} = a_{2} + jb_{2}

Resultant
Voltage V = V_{1} + V_{2} =
(a_{1} + jb_{1}) + (a_{2 }+ jb_{2})

= (a_{1} + a_{2}) + j(b_{1} + b_{2})

Magnitude of resultant, V

Angle from OX-axis,

(ii)Subtraction is done similar to what was done in phasor addition

V = V_{1} −V_{2}
= (a_{1} + jb_{1}) − (a_{2 }+ jb_{2})

= (a_{1} − a_{2}) +
j(b_{1} − b_{2})

Magnitude of resultant voltage,

Angle from OX-axis,

** 9.5
Multiplication and Division of Phasors**

Multiplication and division of phasor is done in polar form as the method is simpler compared to what is done in rectangular form. Consider two phasor:

V_{1} = a_{1} + jb_{1} = V_{1}∠θ_{1}

V_{2} = a_{2} + jb_{2 }= V_{2}∠θ_{2}

**9.5.1 Multiplication**

**(i)**** ****Rectangular
form,**

V_{1} × V_{2} =
(a_{1} + jb_{1}) (a_{2} + jb_{2})

= a_{1}a_{2} + ja_{1}b_{2
}+ ja_{2}b_{1 }+ j^{2}b_{1}b_{2}

= (a_{1}a_{2}−b_{1}b_{2})
+ j(a_{1}b_{2} + a_{2}b_{1})
(as j^{2}=−1)

Magnitude of resultant

Angle w.r.t. OX-axis,

**(ii)**** ****Polar
form**. To multiply the phasors that are in polar form,
multiply their magnitudes and add the angles (algebraically).

V_{1} × V_{2} = V_{1}∠θ_{1}×
V_{2} ∠θ_{2} = V_{1}V_{2}∠
θ_{1}+ θ_{2}

Multiplication of phasors becomes easier when they are expressed in polar form.

**9.5.2 Division**

**(i)**** **** Rectangular
form,**

**(ii)**** ****Polar Form**. To
divide the phasors that are in polar form, the magnitude of phasors are divided
and denominator angle is subtracted from the numerator angle.

**9.6
Transformation **

**9.6.1
Star to delta (Y/∆) transformation**

In any electrical system a star Y connection may be
replaced by an equivalent ∆-connected system. A 3-phase star system
having voltage *V _{L}* and line current I

**Fig.
9.2 Star to delta (Y/∆) transformation**

For a balanced star-connected load, let

*V _{L}* = line voltage;

I_{L} = line
current;

Z∠∅ = impedance per phase

Then for an equivalent ∆-connected system,

Phase voltage *V _{ph}* =

Phase current *I _{ph}*

*Z _{ph}*
= 3Z∠∅

*Z _{Y}*
=

** 9.6.2 Delta to star (∆/Y)
transformation**

**Fig. 9.3 Delta to star (∆/Y)
transformation**

Now, in the equivalent ∆-connected systems, the line voltages and currents must have the same values as in the Y-connected system, hence we must have

*V _{L}*=

*I _{ph}*

*Z _{∆}*
=

*Z _{∆}*
∠ ∅ = 3

*Z _{∆}*
= 3

**Numerical**

**1. A 220
V 3 phase Y connected load is shown in following diagram. The phase sequence is
RYB. Calculate the line currents and neutral line current.**

Z_{R} = (20 + j10) = 22.36∠26.56^{o}

Z_{Y }= (18 + j6) = 18.97∠18.43^{o}

Z_{Y }= (25 + j5) = 25.49∠11.30^{o}

Let
line voltages with difference of phase angle 120^{o}

V_{RN} = 220∠0^{o}

V_{YN} = 220∠─120^{o}

V_{BN} = 220∠─240^{o}

**Line
currents:**

I_{R}
= V_{RN}/Z_{R} = =9.83∠─26.56^{o} = 8.79 ─ j4.39 A

I_{Y}
= V_{YN}/Z_{Y} = =11.59∠─138.43^{o} = ─ 8.67 ─ j7.69 A

I_{B}
= V_{BN}/Z_{B} = =8.63∠─251.3^{o} = ─ 2.76 + j8.17 A

I_{N}
= I_{R} + I_{Y}^{ }+ I_{B}

= 8.79 ─ j4.39 ─ 8.67 ─ j7.69 ─ 2.76 + j8.17

I_{N} = ─ 2.64 + j4.87
= 5.53∠118.46^{o} A

**2. ****Consider 75
ohm resistors are connected in star and then in delta. If line voltage V _{L}
= 440 V calculate line and phase current in star and delta system.**

**Solution**

**Star connection**

Phase voltage *V _{ph}* =

Phase current *I _{ph}*

Line current I_{L} = Phase current I_{ph}
= 3.38 A

**Delta connection**

Phase voltage *V _{ph}* = Line voltage

Phase current *I _{ph}*

Line current *I _{L}*