SULIT
UNIVERSITI MALAYSIA PERLIS
TEST# 1
30 August 2010
DKT 213 – Electrical Technology
Time: 09:00 – 10.00 p.m
Please make sure that this question paper has EIGHT (8) printed pages including this front
page before you start the examination.
This question paper has TWO (2) questions. Answer ALL questions in this paper.
Q1
Q2
31
34
Total
65
%
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(DKT213)
-2Question 1
(a)
Draw (in Figure 1a) how to make an iron nail become a simple electromagnet.
[3 marks]
Figure 1a
(b)
Explain how to make an electromagnet stronger.
[3 marks]
i)
ii)
iii)
(c)
Wrap the coil around an iron core
Add more turns to the coil
Increase the current flowing through the coil.
A wire is shown in Figure 1b which is carrying 5.0 A in the presence of a
magnetic field. Calculate the magnitude and direction of the force induced on the
wire.
Figure 1b
[3 marks]
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(d)
(DKT213)
-3A core with three legs is shown in Figure 1c. Its depth is 5 cm, and there are 200
turns on the leftmost leg. The relative permeability of the core can be assumed to
be 1500 and constant. Assume a 4% increase in the effective area of the air gap
due to fringing effects.
i)
Figure 1c
Determine the reluctance ℜ1 , ℜ 2 , ℜ 3 , ℜ 4 and total reluctance, ℜ total of the
core.
This core can be divided up into four regions. Let R1 be the reluctance of
the left-hand portion of the core, R2 be the reluctance of the center leg of
the core, R3 be the reluctance of the center air gap, and R4 be the
reluctance of the right-hand portion of the core. Then the total reluctance
of the core is
The total reluctance is
[10 marks]
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-4ii)
Determine the flux,Ø in each of the three legs of the core?
[6 marks]
The total flux in the core is equal to the flux in the left leg:
The fluxes in the center and right legs can be found by the “flux divider
rule”, which is analogous to the current divider rule.
iii)
What is the flux density, B in each of the legs?
[6 marks]
The flux density in the legs can be determined from the equation Ø = BA:
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-5-
Question 2
(a)
The secondary winding of a transformer has a terminal voltage of
vs(t) = 282.8 sin 377V. The turns ratio of the transformer is 100:200 (a = 0.50). If
the secondary current of the transformer is is(t) 7.07 sin (377t- 36.87°) A.
The impedances of this transformer referred to the primary side are:
Req= 0.20 Ω
Xeq= 0.75 Ω
i)
Rc= 300 Ω
Xm= 80 Ω
Determine the primary current of this transformer?
[8 marks]
Since no particular equivalent circuit was specified, we are using the
approximate equivalent circuit referred to the primary side.
The secondary voltage and current are
The secondary voltage referred to the primary side is
The secondary current referred to the primary side is
The primary circuit voltage is given by
The excitation current of this transformer is
Therefore, the total primary current of this transformer is
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-6-
ii)
What are its voltage regulation and efficiency?
[6 marks]
The voltage regulation of the transformer at this load is
The input power to this transformer is
The output power from this transformer is
Therefore, the transformer’s efficiency is
(b)
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 = 230 V
Ioc = 0.45 A
Poc = 30 W
Short-circuit test
Vsc = 19.1 V
Isc = 8.7 A
Psc = 42.3 W
All data given were taken from the primary side of the transformer.
i)
Find the equivalent circuit of this transformer referred to the low-voltage side of the transformer.
[12 marks]
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-7-
To convert the equivalent circuit to the secondary side, divide each impedance by the square of the
turns ratio (a = 230/115 = 2). The resulting equivalent circuit is shown below:
ii) Find the transformer’s voltage regulation at rated conditions and 0.8 PF lagging
[4 marks]
To find the required voltage regulation, we will use the equivalent circuit of the transformer
referred to the secondary side. The rated secondary current is
We will now calculate the primary voltage referred to the secondary side and use the voltage
regulation equation for each power factor.
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-8iii) Determine the transformer’s efficiency at rated conditions and 0.8 PF lagging.
[4 marks]
At rated conditions and 0.8 PF lagging, the output power of this transformer is
The copper and core losses of this transformer are
Therefore the efficiency of this transformer at these conditions is
-ooOoo-
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