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BACHELOR OF TECHNOLOGY IN ELECTRONICS AND COMPUTER ENGINEERING
EEE 2213: ELECTRICAL MACHINES I
CAT 2
DATE: 4/05/2021

Time: 6:00 PM-8
...
With the aid of a typical ac electric supply scheme, explain why the Transformers have
made the electric system almost universally AC
(3 marks)
Solution
The electric power is generated at relatively low voltages (e
...
, 11 kV) which then is raised to
very high voltages by means of a transformer and then transmitted
...
Geographically close
to the use points, the electric power is transformed back to safe low utility voltages (400/220
V) (1 mark)
...
1
...


Fig Q
...
c shows a typical ac electric supply scheme
...
A 10  resistance is connected across the secondary winding of a single-phase power
transformer whose secondary voltage is 240 V
...
8 or 5 or 24:5
(1 mark)
2

1

3
...
The primary is connected to 5kV, 50-Hz supply
...
𝐼2 = 𝑎𝐼1 = 10𝑥5 =

50𝐴
ii The secondary emf
𝑃𝑟𝑖𝑣𝑎𝑟𝑦 𝑣𝑜𝑙𝑡𝑎𝑔𝑒,𝐸

e
...
f
...
𝑜𝑓 𝑡𝑢𝑟𝑛𝑠 𝑜𝑛 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 1𝑠𝑖𝑑𝑒,𝑁 =
1

5𝑥103
300

= 16
...
m
...
, 𝐸2 = e
...
f
...
𝑜𝑓 𝑡𝑢𝑟𝑛𝑠 𝑜𝑛 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑠𝑖𝑑𝑒, 𝑁2 = 16
...
44𝑓𝑁1 Ф𝑚 ; ⇒ Ф𝑚 =

𝐸1
5000
=
= 0
...
44𝑓𝑁1
4
...
1𝑚𝑊𝑏
4
...

(3 marks)
Solution

The connection diagram for short circuit test or impedance test on transformer is as shown in the
figure below
...
Voltage is applied to the HV side
and increased from the zero until the ammeter reading equals the rated current
...

The ammeter reading gives primary equivalent of full load current (Isc)
...
Hence, core
loss due to small applied voltage can be neglected
...

2
Therefore,𝑊 = 𝐼𝑆𝐶
𝑅𝑒𝑞 (where 𝑅𝑒𝑞 is the equivalent resistance of transformer)
...

Hence, it is seen that the short circuit test gives copper losses of transformer and approximate
equivalent resistance and reactance of the transformer
...
Highlight characteristic of DC shunt motor and list Four of its applications (3 marks)
Solution

Application

6
...
005Ω
𝑋𝑝 = 6Ω
𝑋𝑆 = 0
...

i Find the equivalent circuit of this transformer referred to the low-voltage side
...
85 PF
lagging
...
88
...
Therefore, the base
impedance on the primary side is

Since𝑍𝑝𝑢 = 𝑍𝑎𝑐𝑡𝑢𝑎𝑙 ⁄𝑍𝑏𝑎𝑠𝑒 , the resulting per-unit equivalent circuit is as shown below:

ii Assume that this transformer is supplying rated load at 277 V and 0
...
What is this transformer’s input voltage? What is its voltage
regulation?
To simplify the calculations, use the simplified equivalent circuit referred to the secondary side
of the transformer:

The secondary current in this transformer is

Therefore, the primary voltage on this transformer (referred to the secondary side) is

The voltage regulation of the transformer under these conditions is

iii What are the copper losses and core losses in this transformer under the
conditions of part (iii)?

iv What is the transformer’s efficiency under the conditions of part (iii)?

7
...
The primary
and secondary resistances are 0
...
01Ω respectively and the corresponding
leakage reactance are 0
...
025Ω respectively
...
1 + 0
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46 
...
6 + 0
...
𝟒𝟔2 + 𝟏
...
568948  or 1
...
46

𝑐𝑜𝑠 𝜑𝑒 = 𝑍𝑒 = 1
...
46

and phase angle of impedance, 𝜑𝑒 = 𝑐𝑜𝑠 −1 (1
...
95

600 2
100

) = 1
...
59 PM

a) A 75kW, 250V DC shunt machine has an armature and field resistances of 0
...
Determine the total armature power developed when working
i as generator delivering 75kW output
ii as motor taking 75kW input
b) A 8-pole, 64 conductor, lap-wound DC shunt generator with terminal voltage of 250V
delivering 10A to the load has Ra=3 Ω and field circuit resistance of 220Ω
...

i Calculate the flux per pole in the machine
ii If the machine has to be run as a motor with the same terminal voltage and
drawing 8A from the mains, maintaining the same magnetic field, find the speed
of the machine
...

d) Derive the EMF equation of DC generator and motor
e) An 4-pole, 25-kW, 120-V dc generator has a duplex lap-wound armature which has 32
coils with 10 turns per coil
...

i
...

What is the current per path in the armature of this generator at the rated load?
iii
...
2Ω and field resistance of 95Ω
...
5A
...
Determine
i the speed regulation,

ii line current and power input when the motor delivers rated voltage
...
05Ω, 0
...

Calculate the generated voltage and the armature current
...

h) Highlight characteristic of dc series motor and list Four of its applications
i) Write brief notes on Brake test, Swinburne’s test, and Regenerative/Hopkinson’s test of
DC machines including their merits and demerits
...
A 4-kVA, 200/400 V, single-phase transformer takes 0
...
When
the low-voltage winding is short-circuited and 15 V is applied to the high-voltage terminals, the
current and power are 10 A and 75 W respectively
...
80 power-factor lagging
...
Derive the condition for maximum efficiency of a transformer
...


3
...
A 10  resistance is connected across the secondary winding of a single-phase power
transformer whose secondary voltage is 120 V
...
A 50 KVA transformer has 600 turns on the primary and 30 turns on the secondary
winding
...
Neglecting the leakage drop,
find
i The primary and secondary currents
ii The secondary emf
iii The maximum flux in the core
6
...
9𝑉 exclusive of brushes
𝑉𝐹 = 440𝑉
𝐼𝐴 = 500𝐴
𝐼𝐹 = 7
...
1𝐴

𝐼𝐹 = 7
...
]

7
...

8
...

The results of the tests are shown below
...

ii Find the impedances of the approximate equivalent circuit referred to the highvoltage side, and sketch that circuit
iii Find the transformer’s voltage regulation at rated conditions and (1) 0
...
0 PF, (3) 0
...

iv Determine the transformer’s efficiency at rated conditions and 0
...

9
...
03Ω
𝑅2 = 0
...
092Ω
𝑋2 = 1
...
8 lagging
...

10
...
The
impedance of the transmission line is 20 + j60 Ω, and the impedance of the load is 10+j5
Ω
...

i
Find the base voltage, current, impedance, and apparent power at every point in
the power system
...

iii Find the power supplied to the load in this system
...

11
...
005Ω
𝑋𝑝 = 5Ω
𝑋𝑆 = 0
...

i Find the equivalent circuit of this transformer referred to the low-voltage side
...
80 PF
lagging

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