Download - EM-Problems Second Semester 2018-2019
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EM-Problems Second Semester 2018-2019
Problem-1
A transformer has 500 primary turns and 3000 secondary turns. If the primary voltage is 240 V,
determine the Secondary voltage, assuming an ideal transformer.
Solution:
For an ideal transformer, voltage ratio = turns ratio, i.e.
π1
π2=
π1
π2, βππππ
240
π2=
500
3000
πβπ’π π ππππππππ¦ π£πππ‘πππ π2 = (3000) (240)
(500)= 1440 π ππ 1.44 ππ
Problem-2
An ideal transformer with a turns ratio of 2:7 is fed from a 240 V supply. Determine its output
voltage.
Solution:
A turns ratio of 2:7 means that the transformer has 2 turns on the primary for every 7 turns on the
secondary (i.e. a step β up transformer). Thus,
π1
π2=
2
7
πΉππ ππ πππππ π‘ππππ ππππππ,π1
π2=
π1
π2 ; βππππ
2
7=
240
π2
πβπ’π π‘βπ π ππππππππ¦ π£πππ‘πππ π2 = (240) (7)
(2)= 840 π
Problem-3
An ideal transformer has a turns ratio of 8:1 and the Primary current 3 A when it is supplied at 240
V. Calculate the secondary voltage and current.
Solution:
π¨ πππππ πππππ ππ π: π πππππ π΅π
π΅π=
π
π, π. π. π ππππ β π πππ πππππππππππ.
π΅π
π΅π=
π½π
π½π ππ ππππππ πππ πππππππ π½π = π½π (
π΅π
π΅π) = πππ (
π
π) = ππ π½ππππ
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Problem-4
An ideal transformer, connected to 240 V mains, supplies a 12 V, 150 W lamp. Calculate the
transformer turns ratio and the current taken from the supply.
Solution:
π1 = 240π, π2 = 12π, πΌ2 = π
π2=
150
12= 12.5 π΄
ππ’πππ πππ‘ππ = π1
π2=
π1
π2=
240
12= 20
π1
π2=
πΌ2
πΌ1, ππππ π€βππβ, πΌ1 = πΌ2 (
π2
π1) = 12.5 (
12
240)
π»ππππ ππ’πππππ‘ π‘ππππ ππππ π‘βπ π π’ππππ¦, πΌ1 = 12.5
20= 0.625 π΄
Problem-5
A 5-kVA single-phase transformer has a turns ratio of 10:1 and is fed from a 2.5 kV supply.
Neglecting losses, determine:
a) the full-load secondary current.
b) the minimum load resistance which can be connected a cross the secondary winding to give
full load kVA.
c) the primary current at full load kVA.
Solution: π1
π2=
10
1 πππ π1 = 2.5 ππ = 2500 π
πππππ π1
π2=
π1
π2, π ππππππππ¦ π£πππ‘πππ
π2 = π1 (π2
π1) = 2500 (
1
10) = 250 π
πβπ π‘ππππ ππππππ πππ‘πππ ππ π£πππ‘ β πππππππ = π2 πΌ2 (ππ‘ ππ’ππ ππππ), π. π. 5000 = 250 πΌ2
π»ππππ ππ’ππ ππππ π ππππππππ¦ ππ’πππππ‘ πΌ2 = 5000
250= 20 π΄
ππππππ’π π£πππ’π ππ ππππ πππ ππ π‘ππππ, π πΏ = π2
πΌ2=
250
20= 12.5Ξ©
π1
π2=
πΌ2
πΌ1, ππππ π€βππβ πππππππ¦ ππ’πππππ‘, πΌ1 = πΌ2 (
π2
π1) = 20 (
1
10) = 2 π΄
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Problem-6
A 2400 V/400 V single-phase transformer takes a no-load current of 0.5 A and the core
losses 400 W.
Determine the values of the magnetizing and core loss components of the no-load current
(Io or Ie). Draw to scale the no-load phasor diagram for the transformer.
Solution:
π1 = 2400 π, π2 = 400 π, πΌπ = 0.5 π΄
The no-load phasor diagram is shown in fig. below
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Problem-7
A transformer takes a current of 0.8 A when its primary is connected to a 240 volt, 50 Hz supply, the
secondary being on open circuit. If the power absorbed is 72 watts, determine
a) The iron loss current.
b) The power factor on no-load.
c) The magnetizing current.
Solution:
Io = 0.8 A, V = 240 V
a) Power absorbed = total core loss = 72 Watt = V1 Io cos Οo
Hence 72 = 240 Io cos Οo
And iron loss current, πΌπ = πΌπ cos Ξ¦π = 72
240= 0.3 π΄
πππ€ππ ππππ‘ππ ππ‘ ππ ππππ, cos Ξ¦π = πΌπ
πΌπ=
0.3
0.8= 0.375
πΌπ2 = πΌπ
2 + πΌπ2
πΌπ = β(πΌπ2 β πΌπ
2)
πΌπ = β(0.82 β 0.32)
πΌπ = 0.74 π΄
Π€m
IC
IM
IO = Ie
E2
E1
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Problem-8 A 100 kVA, 4000 V/200 V, 50 Hz single-phase transformer has100 secondary turns. Determine
a) The primary and secondary current.
b) The number of primary turns.
c) The maximum value of the flux.
Note:
The magnetic flux set up in the core of a transformer when an alternating voltage is applied to its
primary winding is also alternating and is sinusoidal.
Let ΙΈm be the maximum value of the flux
and f be the frequency of the supply. The
time for 1 cycle of the alternating flux is
the periodic time T, where T = 1/f
seconds
The flux rises sinusoidally from zero to
its maximum value in ΒΌ cycle, and the
time for ΒΌ cycle is ΒΌ f seconds.
Hence the average rate of change of flux =
ΙΈm / (1/4 f) = 4 f ΙΈm Wb/S and since 1Wb/s
= 1volt, the average e.m.f. induced in each turn = 4 f ΙΈm volts.
As the flux varies sinusoidally, then a sinusoidal e.m.f. will be induced in each turn of both primary
and secondary windings.
For a sine wave, form factor = rms value/average value = 1.11
Hence rms value = form factor * average value = 1.11 * average value
Thus rms e.m.f. induced in each turn = 1.11 * 4 f ΙΈm volts = 4.44 f ΙΈm volts
Therefore, rms value of e.m.f. induced in primary,
E1 = 4.44 f ΙΈm N1 volts
And rms value of e.m.f. induced in secondary,
E2 = 4.44 f ΙΈm N2 volts
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Solution:
V1 = 400V, V2 = 200 V, f = 50 Hz, N2 = 100 turns
a) Transformer rating = V1 * I1 = V2 *I2 = 100000 VA
hence primary current, πΌ1 = 100000
π1=
100000
4000= 25 π΄
and secondary Current, πΌ2 = 100000
π2=
100000
200= 500 π΄
from which, primary turns, π1 = (π1
π2) (π2) = (
4000
200) (100)
b) i.e., N1 = 2000 turns
c)
ππππ π€βππβ, πππ₯ππππ’π πππ’π₯ Ξ¦π = πΈ2
4.44 π π2=
200
4.44 (50) (100)
(assuming E2 = V2)
ππ = 9.01 β 10β3 ππ ππ 9.01 πππ
Where E1 = 4.44 f Ξ¦m N1
From which, ππ = πΈ1
4.44 π π1=
4000
4.44 (50) (2000)
(assuming E1 = V1)
ππ = 9.01 β 10β3 ππ ππ 9.01 πππ
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Problem-9
A single-phase, 50 Hz transformer has 25 primary turns and 300 secondary turns. The cross-
sectional area of the core is 300 cm2. When the primary winding is connected to a 250 V supply,
determine
a) The maximum value of the flux density in the core.
b) The voltage induced in the secondary winding.
Solution:
e.m.f. E1 = 4.44 f ΙΈm N1 volts i.e.,
250 = 4.44 (50) Ξ¦m (25)
ππππ π€βππβ, πππ₯πππ’π πππ’π₯ ππππ ππ‘π¦, ππ = 250
(4.44) (50) (25) ππ = 0.04505 ππ
However,ππ = π΅π β π΄, π€βπππ π΅π = πππ₯πππ’π πππ’π₯ ππππ ππ‘π¦ ππ π‘βπ ππππ πππ
π΄ = ππππ π β π πππ‘πππππ ππππ ππ π‘βπ ππππ.
Hence Bm * 300 * 10-4
= 0.04505
From which, maximum flus density, π΅π = 0.04504
300β 10β4 = 1.5 π
b) π1
π2=
π1
π2, ππππ π€βππβ, π2 = π1 (
π2
π1)
π. π. , π£πππ‘πππ ππππ’πππ ππ π‘βπ π ππππππππ¦ π€ππππππ,
π2 = (250) (300
25) = 3000 π ππ 3 ππ
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Problem-10
A single-phase 500 V/100 V, 50 Hz transformer has a maximum core flux density of 1.5 T and an
effective core cross-sectional area of 50 cm2. Determine the number of primary and secondary turns.
Solution:
The e.m.f. Equation for a transformer is E = 4.44 f ΙΈm N
And maximum flux, ππ = π΅ β π΄ = (1.5) (50 β 10β4) = 75 β 10β4 ππ
Since E1 = 4.44 f ΙΈm N1
Then primary turns,π1 = πΈ1
4.44 π β π=
500
4.44 (50) (75β 10β4)= 300 π‘π’πππ
Since E2 = 4.44 f ΙΈm N2
Then secondary turns, π2 = πΈ2
4.44 π β π=
500
4.44 (50) (75β 10β4)= 60 π‘π’πππ
Problem-11
A 4500 V/225 V, 50 Hz single-phase transformer is to have an approximate e.m.f. per turn of 15 V
and operate with a maximum flux of 1.4 T. Calculate
a) The number of primary and secondary turns.
b) The cross-sectional area of the core.
Solution:
a) πΈ. π. π. πππ π‘π’ππ = πΈ1
π1=
πΈ2
π2= 15
π»ππππ πππππππ¦ π‘π’πππ , π1 = πΈ1
15=
4500
15= 300
πππ π ππππππππ¦ π‘π’πππ , π2 = πΈ2
15=
225
15= 15
b) E.m.f. E1 = 4.44 f ΙΈm N1
from which,Ξ¦π = πΈ1
4.44 π π1=
4500
4.44 (50) (300)= 0.0676 ππ
πππ€ πππ’π₯ ππ = π΅π β π΄, π€βπππ π΄ ππ π‘βπ ππππ π β ππππ‘πππππ ππππ ππ π‘βπ ππππ, βππππ
ππππ π΄ = ππ
π΅π=
0.0676
1.4= 0.0483 π2 ππ 483 ππ2
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Problem-12
A single-phase transformer has 2000 turns on the primary and 800 turns on the secondary. Its no
load current is 5 A at a power factor of 0.20 lagging. Assuming the volt drop in the windings is
negligible, determine the primary current and power factor when the secondary current is 100 A at a
power factor of 0.85 lagging.
Solution:
Let π°πβ² be the component of the primary current which provides the
restoring mmf. Then
πΌ1β² π1 = πΌ2 π2
i.e.,
πΌ1β² (2000) = (100) (800)
ππππ π€βππβ, πΌ1β² =
(100) (800)
2000= 40 π΄
If the power factor of the secondary is 0.85 the cos Ο2 = 0.85, from which, Ο2 = arccos 0.85 = 31.8o.
If the power factor on no load is 0.2, then cos Ο0 = 0.2, and Ο0 = arccos 0.2 = 78.5o. In the phasor
diagram shown in fig. above I2 = 100 A is shown at an angle of Ο2 = 31.8o
to V2 and π°πβ² = 40 A is
shown in anti-phase to I2
The no load current I0 = 5 A is shown at an angle of Ο0 = 78.5o to V1.
Current I1 is the phasor sum of π°πβ² and I0 and by drawing to scale, I1 = 44 A and angle Ο1= 37
o
π΅π¦ πππππ’πππ‘πππ, πΌ1 cos π1 = ππ + ππ = πΌπ cos ππ + πΌ1β² cos π2
= (5) β (0.2) + (40) β (0.85) = ππ. π π¨
πππ πΌ1 sin π1 = ππ + ππ = πΌπ sin ππ + πΌ1β² sin π2
= (5) sin 78.5π + (40) sin 31.8 π = ππ. ππ π¨
π»ππππ π‘βπ ππππππ‘π’ππ ππ πΌ1 = β(35.02 + 25.982) = ππ. ππ π¨
πππ tan π1 = (25.98
35.0) , ππππ π€βππβ,
π1 = arctan (25.98
35.0) = ππ. πππ
π»ππππ π‘βπ πππ€ππ ππππ‘ππ ππ π‘βπ πππππππ¦ = cos π1 = cos 36.59π = π. π π
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Problem-13
A transformer has 600 primary turns and 150 secondary turns. The primary and secondary
resistances are 0.25 Ξ© and 0.01 Ξ© respectively and the corresponding leakage reactances are 1.0 Ξ©
and 0.04 Ξ© respectively. Determine:
a) The equivalent resistance referred to the primary winding
b) The equivalent reactance referred to the primary winding
c) The equivalent impedance referred to the primary winding
d) The phase angle of the impedance.
Solution:
π ππ = π 1 + π 2 (π1
π1)
2
π ππ = 0.25 + 0.01 (600
150)
2
= 0.41 Ξ©
πππ = π1 + π2 (π1
π1)
2
π ππ = 1.0 + 0.04 (600
150)
2
= 1.64 Ξ©
ππ = β(π π2 + ππ
2) = 1.69 Ξ©
ππ = π΄πππππ (0.41
1.69) = 75.96π
Problem-14
A 5 kVA, 200V/400V, single-phase transformer has a secondary terminal voltage of 387.6 volts
when loaded. Determine the regulation of the transformer.
Solution:
ππ = ππ ππππ πππππππππ¦ ππππ‘πππ β π‘πππππππ π£πππ‘πππ ππ ππππ
ππ ππππ πππππππππ¦ ππππ‘πππ β 100%
ππ = 400 β 387.6
400 β 100% = 3.1 %
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Problem-15
The equivalent circuit impedances of a 20-kVA, 8000/240 V, 60-Hz transformer are to be
determined. The open-circuit test was performed on the secondary side of the transformer (to reduce
the maximum voltage to be measured) and the short circuit test were performed on the primary side
of the transformer (to reduce the maximum current to be measured). The following data were taken:
Open β Circuit test
(on secondary)
Short β Circuit test
(on Primary)
VOC = 240 V VSC = 489 V
IOC = 7.133 A ISC = 2.5 A
POC = 400 W PSC = 240 W
Find the impedances of the approximate equivalent circuit referred to the primary side, and
sketch that circuit.
Solution:
The turns ratio of this transformer is a = 8000/240 = 33.3333. The power factor during the open-
circuit test is
Therefore, the values of the excitation branch referred to the low-voltage (secondary) side are
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