2. In last class, we have seen about the balanced
and unbalanced loads of three phase circuit.
Depending upon the impedance matching the
loads are being classifies as balanced and
unbalanced loads.
These loads has their own power equation
and other three phase quantities.
Today we will see about the star delta
conversion.
7. Advantages
1. The primary side is star connected. Hence fewer
number of turns are required. This makes the
connection economical
2. The neutral available on the primary can be
earthed to avoid distortion.
3. Large unbalanced loads can be handled
satisfactory.
7
8. Disadvantages
The secondary voltage is not in phase with the
primary. (30 ⁰ phase difference )
Hence it is not possible to operate this connection
in parallel with star-star or delta-delta connected
transformer.
8
9. Wye(star) to Delta Transformation:
Consider the following:
a
bc
a
bc
Ra
RbRc
R1 R2
R3
(a) wye configuration (b) delta configuration
a
accbba
c
accbba
b
accbba
R
RRRRRR
R
R
RRRRRR
R
R
RRRRRR
R
3
2
1
321
31
321
32
321
21
RRR
RR
R
RRR
RR
R
RRR
RR
R
c
b
a
10. Using the following circuit. Find Req.
9
10 5
8 4
V
+
_
Req 10
I
a
bc
Convert the delta around a – b – c to a wye.
13. Features
secondary Phase voltage is 1/√3 times of line
voltage
neutral in secondary can be grounded for 3 phase
4 wire system
Neutral shifting and 3rd harmonics are there
Phase shift of 30⁰ between secondary and primary
currents and voltages
13
14. The three-phase ac systems are considered as a
balanced circuit, made up of a balanced three-
phase source, a balanced line, and a balanced
three-phase load.
The star-delta (Y-Δ) or delta-star (Δ-Y)
conversion is required in three-phase ac
systems to simplify the circuits and ease their
analysis.
If a three-phase supply or a three-phase load is
connected in delta, it can be transformed into an
equivalent star-connected supply or load. After
the analysis, the results are converted back into
their original delta equivalent.
Hinweis der Redaktion
In Delta:
Equivalent Resistance between A & B
="Rab in parallel with (Rbc+ Rca)"= (𝑅_𝑎𝑏 (𝑅_𝑏𝑐+𝑅_𝑐𝑎))/(𝑅_𝑎𝑏+𝑅_𝑏𝑐+𝑅_𝑐𝑎 )
In Star:
Equivalent Resistance between A & B
〖=𝑅〗_𝑎+〖 𝑅〗_𝑏
Therefore,
𝑅_𝑎+〖 𝑅〗_𝑏=(𝑅_𝑎𝑏 (𝑅_𝑏𝑐+𝑅_𝑐𝑎))/(𝑅_𝑎𝑏+𝑅_𝑏𝑐+𝑅_𝑐𝑎 )
Similarly,
𝑅_𝑏+〖 𝑅〗_𝑐=(𝑅_𝑏𝑐 (𝑅_𝑐𝑎+𝑅_𝑎𝑏))/(𝑅_𝑎𝑏+𝑅_𝑏𝑐+𝑅_𝑐𝑎 )
And
𝑅_𝑐+〖 𝑅〗_𝑎=(𝑅_𝑐𝑎 (𝑅_𝑎𝑏+𝑅_𝑏𝑐))/(𝑅_𝑎𝑏+𝑅_𝑏𝑐+𝑅_𝑐𝑎 )
On adding any two of the above equations and subtracting with the 3rd one gives
𝑅_𝑎=(𝑅_𝑎𝑏 𝑅_𝑐𝑎)/(𝑅_𝑎𝑏+𝑅_𝑏𝑐+𝑅_𝑐𝑎 )
𝑅_𝑏=(𝑅_𝑏𝑐 𝑅_𝑎𝑏)/(𝑅_𝑎𝑏+𝑅_𝑏𝑐+𝑅_𝑐𝑎 )
𝑅_𝑐=(𝑅_𝑐𝑎 𝑅_𝑏𝑐)/(𝑅_𝑎𝑏+𝑅_𝑏𝑐+𝑅_𝑐𝑎 )