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Basic Electrical Technology (DET 211/3)
Tutorial Module
TUTORIAL 4 – SINGLE – PHASE TRANSFORMER
Question 1:
The secondary winding of a transformer has a terminal voltage of vs (t) = 282.8
sin 377t V. The turns ratio of the transformer is 100:200. If the secondary current
of the transformer is (t) =7.07 sin (377t - 36.87o) A, what is the primary current of
this transformer? What are its voltage regulation and efficiency? The impedances
of this transformer referred to the primary side are
Req = 0.20Ω
Rc = 300Ω
Xeq = 0.750Ω
XM = 80Ω
(Ans: I p  11.1  41o A , VR = 6.2%, η = 93.7%)
Question 2:
A 20-kVA 8000/480-V distribution transformer has the following resistances and
reactances:
Rp = 32Ω
Rs = 0.05Ω
Xp = 45Ω
Xs = 0.06Ω
RC = 250kΩ XM = 30kΩ
The excitation branch impedances are given referred to the high-voltage side of
the transformer.
a) Find the equivalent circuit of this transformer referred to the high-voltage
side.
b) Find the per-unit equivalent circuit of this transformer.
c) Assume that this transformer is supplying rated load at 480V and 0.8 PF
lagging. What is this transformer’s input voltage? What is its voltage
regulation?
d) What is the transformer’s efficiency under the conditions of part (c)?
(Ans: a) Rs’ = 13.9Ω, Xs’ = 16.7Ω b) Rp = 0.01, Xp = j0.0141, Rs’ = 0.0043, Xs’ =
j0.0052, RC = 78.125, XC = j9.375 c) V p  81850.38o V , VR = 2.31% d) η =
96.6%)
UNIVERSITI MALAYSIA PERLIS (UniMAP) –Tutorial 4
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Basic Electrical Technology (DET 211/3)
Tutorial Module
Question 3:
A 1000-VA 230/115-V transformer has been tested to determine its equivalent
circuit. The results of the tests are shown below.
Open circuit test
VOC = 230V
IOC = 0.45A
POC = 30W
Short circuit test
VSC = 19.1V
ISC = 8.7A
PSC = 42.3W
All data given were taken from the primary side of the transformer.
a) Find the equivalent circuit of this transformer referred to the low-voltage
side of the transformer.
b) Find the transformer’s voltage regulation at rated conditions and (1) 0.8
PF lagging, (2) 1.0 PF, (3) 0.8 PF leading.
c) Determine the transformer’s efficiency at rated conditions and 0.8 PF
lagging.
(Ans: a) Req,s =0.140Ω, Xeq,s = j0.532Ω, RC,s =441Ω, XM,s = 134Ω b) VR = 3.3%,
1.1%, -1.5%, η = 94.9%)
Question 4:
A 15-kVA 8000/230V distribution transformer has an impedance referred to the
primary of 80 + j300 Ω. The components of the excitation branch referred to the
primary side are RC = 350kΩ and XM = 70kΩ.
a) If the primary voltage is 7967 V and the load impedance is ZL = 3.0 +
j1.5Ω, what is the secondary voltage of the transformer? What is the
voltage regulation of the transformer?
b) If the load is disconnected and a capacitor of –j4.0Ω is connected in its
place, what is the secondary voltage of the transformer? What is its
voltage regulation under these conditions?
(Ans: Vs  218.8  3.1o V , VR = 4.7% b) Vs  246.5  1.2 o V , VR = -7.07%)
UNIVERSITI MALAYSIA PERLIS (UniMAP) –Tutorial 4
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Basic Electrical Technology (DET 211/3)
Tutorial Module
Question 5:
A 5000-kVA 230/13.8-kV single-phase power transformer has a per-unit
resistance of 1 percent and a per unit reactance of 5 percent (data taken from the
transformer’s nameplate). The open-circuit test performed on the low-voltage
side of the transformer yielded the following data:
VOC = 138kV IOC = 15.1A POC = 44.9kW
a) Find the equivalent circuit referred to the low-voltage side of this
transformer.
b) If the voltage on the secondary side is 13.8 kV and the power supplied is
4000 kW at 0.8 PF lagging, find the voltage regulation of the transformer.
Find its efficiency.
(Ans: a) Req = 0.39Ω, Xeq = 1.9Ω b) VR = 3.84%, η = 97.6%)
UNIVERSITI MALAYSIA PERLIS (UniMAP) –Tutorial 4
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