Example 4.2
A balanced (a b c sequence) Y- connected source with (EAN=100∠100 V)
is connected to a Δ-connected balanced load (8+j4)Ω per phase. Calculate
the phase and line currents.
Solution: This can be solved in two ways.
Method 1: the load impedance is
Z Δ = 8 + j 4 = 8.944∠26.57 0 Ω
If the phase voltage EAN=100∠100, the line voltage is
E AB = E AN 3∠30 0 = 100 × 3∠(10 + 30) 0 = Vab
Or
Vab = 173.2∠40 0 V
The phase current are
I ab =
Vab
173.2∠40 0
=
= 19.36∠13.430 A
Z Δ 8.944∠26.57 0
I bc = I ab ∠ − 120 0 = 19.36∠ − 106.57 0 A
I ca = I ab ∠120 0 = 19.36∠133.43 0 A
The line current are
I a = I ab 3∠ − 30 0 = 3 (19.36)∠(13.43 − 30) 0 = 33.53∠ − 16.57 0 A
I b = I a ∠ − 120 0 = 33.53∠ − 136.57 A
I c = I a ∠120 0 = 33.53∠103.43 0 A
Method 2:
Alternatively, using single-phase analysis
Ia =
E AN
100∠10 0
=
= 33.54∠ − 16.57 0 A
0
⎛ Z Δ ⎞ 2.981∠26.57
⎟
⎜
⎝ 3 ⎠
As above, other line currents are obtained using the (a b c) phase
sequence.
111
4.8 Balanced Delta-Wye Connection
Consider the Δ-Y circuit in fig.(4.18), assuming the (a b c)
sequence, the phase voltage of a Delta-connected source are.
E AB = VP ∠0 0 ,
E BC = V P ∠ − 120 0 ,
E CA = V P ∠120 0
There are also the line voltages as well as the phase voltages
a
A
ECA
EAB
n
C
EBC
b
B
c
Fig.(4.18)
We can obtain the line currents in many ways.
1st method
Apply KVL to loop (A a n b B A) is shown in figure, writing.
E AB − Z Y I a + Z Y I b = 0
Or
Z Y (I a − I b ) = E AB = V P ∠0 0
Thus
VP ∠0 0
Ia − Ib =
ZY
But Ib lags Ia by 1200 , since we assumed the (a b c) sequence, that is
I b = I a ∠ − 120 0 , hence
112
I a − I b = I a (1 − 1∠ − 120 0 )
⎛ 1
3⎞
⎟ = I a 3∠30 0
= ⎜⎜1 + + j
⎟
2
2
⎠
⎝
Substituting we get
I a 3∠30 0 =
VP
Ia =
3
VP ∠0 0
ZY
∠ − 30 0
ZY
From this, we obtain the other line currents Ib and Ic using the positive
phase sequence i.e, I b = I a ∠ − 120 0 , I c = I a ∠120 0 . The phase current are
equal to line currents.
2nd method
To obtain the line currents is to replace the delta-connected source
with its equivalent Wye-connected source, as shown in fig.(4.19). We
found that the line-to line voltages of a Wye-connected source lead their
corresponding phase voltage by 300.
E AB = 3E AN ∠30 0 ,
E AN =
E AB
3
∠ − 30 o
Fig.(4.19)
113
Thus, the equivalent Wye-connected source has the phase voltages.
E AN =
E BN =
VP ∠ − 30 0
3
VP ∠ − 150 0
3
,
ECN =
VP ∠90 0
3
The single phase Equivalent
Therefore, we can use the equivalent single-phase circuit as shown in
figure below. From which the line current for phase a is
VP
Ia =
3
∠ − 30 0
ZY
Or, we may transform the Wye- connected load to an equivalent deltaconnected load. This results in a delta-delta system, which can be
analyzed.
114
Example 4-3
A balanced Y-connected load with a phase resistance of 40Ω and a
reactance of 25Ω is supplied by a balanced, positive sequence
Δ-connected source with a line voltage of 210 V, calculate the phase
currents. Use EAB as reference.
Solution:
The load impedance is
Z Y = 40 + j 25 = 47.17∠32 o Ω
And the source voltage is
E AB = 210∠0 o
When the Δ-connected source is transformed to a Y-connected source,
E AN =
E AB
3
∠ − 30 o = 121.2∠ − 30 0 V
The line currents are
Ia =
E AN 121.2∠ − 30 o
=
= 2.57∠ − 62 o A
o
ZY
47.12∠32
I b = I a ∠ − 120 o = 2.57∠ − 182 o A
I c = I a ∠120 o = 2.57∠58 o A
Which are the same as the phase currents.
115
4.9 Balanced Delta- Delta Connection
The source as well as the load may be delta-connected as shown in
fig.(4.20). Our goal is to obtain the phase and line currents as usual.
Assuming a positive sequence, the phase voltages for a delta-connected
source are.
E AB = VP ∠0 0 ,
E BC = V P ∠ − 120 0 ,
E CA = V P ∠120 0
The line voltages are the same as the phase voltages. And assuming there
is no line impedances.
E AB = Vab ,
E BC = Vbc ,
ECA = Vca
Hence, the phase current are
I ab =
Vab E AB
=
,
ZΔ
ZΔ
I bc =
Vbc E BC
=
,
ZΔ
ZΔ
I ca =
Vca ECA
=
ZΔ
ZΔ
a
A
Iab
ECA
EAB
Ica
c
C
EBC
b
B
Fig.(4.20)
116
Ibc
The line current are obtained from the phase currents by applying KCL at
nodes a, b, and c.
I a = I ab − I ca ,
I b = I bc − I ab ,
I c = I ca − I bc
Also, we know that, each line current lags the corresponding phase
current by 300, the magnitude IL of the line current is
3
times the
magnitude IP of the phase current.
I L = 3I P
An alternative way of analyzing the Δ-Δ circuit is to convert both the
source and the load to their Y-equivalents.
Example 4-4:
A balanced Δ- connected load having an impedance (20-j15)Ω is
connected
to
Δ-connected,
positive-sequence
generator
having
EAB=330∠00 V. Calculate the phase currents of the load and the line
currents.
Solution:
The load impedance per phase is
Z Δ = 20 − j15 = 25∠ − 36.87 o Ω
The phase currents are
I ab =
Vab
330∠0 o
=
= 13.2∠36.87 o A
Z Δ 25∠ − 36.87 o
I bc = I ab ∠ − 120 o = 13.2∠ − 83.13 o A
I ca = I ab ∠120 o = 13.2∠156.87 o A
For a delta load, the line current always lags the corresponding phase
current by 300 and has a magnitude 3 times that of the phase current.
117
I a = I ab
(
) (
3∠ − 30 o = 13.2∠36.87 o
)(
3∠ − 30 o
= 22.86∠6.87 o A
I b = I a ∠ − 120 o = 22.86∠ − 113.13 o A
I c = I a ∠120 o = 22.86∠126.87 o A
118
)