# Single phase Transformer Questions Rev2 S2-2023[1]

```Single-Phase Transformer Questions
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Draw the exact equivalent circuit diagram of a transformer under-load with a lagging power factor
and also draw its corresponding phasor diagram.
A non-sinusoidal voltage π = 150π ππ314π‘ − 75π ππ1570π‘ is applied to a 250 π‘π’ππ winding of a single
phase transformer. Determine the flux as a function of time.
A voltage π½ = ππππππππππ is applied to the transformer winding during a no-load test. The resulting
π
current was found to be π = π π¬π’π§ (ππππ − ) π¨. Find the core loss and the rms value of the exciting
π
current.
A 100kVA single-phase transformer has 400 turns on the primary and 80 turns on the secondary.
The primary and secondary resistances are 0.3 β¦ and 0.01 β¦ respectively, and the corresponding
leakage reactances are 1.1 β¦ and 0.035 β¦ respectively. The supply voltage is 2200V. Calculate:
(A) The equivalent impedance referred to the primary circuit (B)The voltage regulation and the
secondary terminal voltage for full load having a power factor of (i) 0.8 lagging and (ii) 0.8 leading.
(C) The percentage resistance and leakage reactance drops of the transformer.
A 10-kVA, 2000/400-V, single-phase transformer at no-load has resistances and leakage
reactances as follows: πΉπ = π. π π; πΉπ = π. π π &amp; πΏπ = ππ π; πΏπ = π. ππ π. If the pf is 0.8 lagging,
determine (A) the voltage regulation and the approximate value of the secondary voltage when the
transformer is exposed to nameplate conditions. (B) Use p. u. to confirm your answers.
The no-load current of a single-phase transformer is 1 π΄ at 0.4 power factor when connected across
a 200 π, 50 π»π§ supply. The number of turns on the primary is twice that of the secondary. Find the
primary current if a load taking 50 π΄ at a lagging power factor of 0.8 is now connected across the
secondary.
A 6600/400 π, single-phase transformer has emf per turn of 15 π and maximum flux density of
1.5 ππ/π2 , Find: (a) Suitable number of primary and secondary turns. (b) Cross-sectional area of
the core.
A 250/500-V, single-phase transformer gave the following test results:
O.C.T: 250 π, 2 π΄, 80 π. S.C.T: 40 π 10 π΄ (Full-load) 120 π. Use a power factor of 0.8
lagging and determine: (A)The constant parameters of the transformer (B) The per unit
secondary voltage and the load angle.
−2.86 0 ). (c)The full-load efficiency.
9
The parameters of a 150 πππ΄, 2400/240 π, 50 π»π§, single-phase transformer are as follows: π1 =
0.2 Ω, π2 = 2 &times; 10−3 Ω, π1 = 0.45 Ω, π2 = 4.5 &times; 10−3 Ω
ππΆ1 = 10 π Ω, ππ1 = 1.6 π Ω, Determine (A) the open circuit data when the LV-side is excited
at rated voltage (B) The short circuit data when the LV-side is short-circuited.
10 A 10 πππ΄, 2000/400 π, 50π»π§, single-phase transformer has the following prameters:
HV side: π = 5.5 Ω,
ππΏ = 18 Ω. LV side:
π = 0.5 Ω,
ππΏ = 0.7 Ω
If the AC source on the H.V. side, determine the full-load secondary voltage in per unit for a
lagging power factor of 80%.
11
A 20-kVA, 4000/400-V, 50-Hz, single phase transformer has the following parameters:
π―π½: πΉ = π π, πΏπ³ = π π, πΉπͺ = ππ ππ, πΏπ΄ = π ππ; π³π½: πΉ = π. ππ π, πΏπ³ = π. ππ π
Determine; (11.1) per unit values of impedance referred to both sides (11.2) Voltage regulation
for a lagging power factor of 0.8.
12
The full-load copper loss on the HV side of 1 300-kVA, 11000/550-V transformer is 1860-W and
1440-W on the LV-side. The reactance referred to the primary is 10-β¦. Calculate: (12.1) π1 &amp; π2
(12.2) πππ»π &amp; ) πππΏπ (12.3) Full-load terminal voltage at a pf 0.8 lagging.
13
A 23-kVA, 2300/230-V, 60-Hz, single-phase transformer with ππ = π + πππ π &amp; ππ = π. ππ +
ππ. ππ π. The transformer is operating at 75% of its rated load and uses 230-V terminal voltage.
Determine the efficiency of the transformer if the load is operated at 0.886 lagging.
14
The primary and secondary windings of a 30 πππ΄, 5000/250 π, 50π»π§, single-phase transformer
has a resistance of 10 Ω and 0.015 Ω respectively. The total reactance referred to the HV side
is 25 Ω. Use a power factor of 0.85 lagging and determine: (a) The percentage regulation of
15
Test data obtained on a 30-kVA, 3000/110-V, single-phase transformer is as follows:
O.C.T: 3000-V 0.5-A
350-W
S.C.T:
150-V
10-A
500-W
Calculate: (15.1) The full-load efficiency of the transformer at 0.8 pf lagging. (15.2) The
efficiency of the transformer at half-load, unity pf. (15.3) The kVA output at which the efficiency
is maximum.
16
The following data refers to a single-phase transformer ratio 60:4; ππ = ππ + πππ π; ππ =
π. ππ + ππ. ππ π; π°π = π. ππ π¨ and leads the flux by 60o. The secondary side delivers 200-A at a
terminal voltage of 500-V with a pf 0.85 lagging. Find: (16.1) The primary voltage and pf (16.2)
The machine efficiency.
17
A 10-kVA, 400/200-V, 50-Hz, single-phase transformer gave the following test results:
O.C.T (H.V open): 200-V 13-A 120-W; S.C.T (LV short circuited): 22-V 30-A 200-W
Determine the equivalent parameters referred to both sides.
18
A 20-kVA, 4000/400-V, single-phase transformer requires only 4.5% of the voltage to circulate
full-load current on the low voltage side. If the core losses are 200-W, calculate (18.1) The
voltage regulation for a load with 0.8 lagging pf. (18.2) Efficiency when supplying 18-kW load.
19
A 120-kVA, 2400/240-V, single-phase transformer has ππ = π. ππ + ππ. π π; ππ = π. ππ +
ππ. ππ π. The transformer is designed to operate at maximum efficiency 0f 70% of its rated load
with 0.8 lagging pf. Determine: (19.1) The base load at maximum efficiency. (19.2) Maximum
efficiency. (19.3) Full-load efficiency. (19.4) RC.
20
A 50-kVA, 6360/240-V transformer is tested on open circuit and short circuited to obtain its
efficiency; The results of the tests are as follows: πΆπͺπ»: π½π = ππππ π½, π°π = π π¨, π·ππ =
π ππΎ πππ πΊπͺπ»: π½ππ = πππ π½, π°π = πππ π¨, π·ππ = π ππΎ. Find: (20.1) The equivalent parameters
of the transformer referred to the high voltage side (20.2) The efficiency of the transformer when
supplying full-load at a pf of 0.8 lagging.
21
The primary and secondary windings of a 30-kVA, 5000/250 –V transformer has resistances of
10 β¦ and 0.015 β¦ respectively. The total reactance referred to the primary is 25 β¦. Calculate:
(21.1) The full-load regulation at pf of 0.8 lagging. (21.2) The secondary voltage for a leading
pf of 0.6. (21.3) The full-load efficiency of the transformer for a lagging pf of 0.75 if the maximum
efficiency occurs at 85% of full-load (21.4) The per unit equivalent resistance.
22
A 10-kVA, 250/100-V, 50-Hz, single-phase transformer gave the following test results:
O.C.T: 250-V
2.6-A
0.3 pf
and
S.C.T: 18-V
100-A
240-W. Find:
(22.1) The per unit impedance. (22.2) The voltage regulation for a leading power factor of 0.78.
(22.3) The full-load efficiency at upf. (22.4) The load at which maximum efficiency occurs. (22.5)
The value of the maximum efficiency at upf.
23
The maximum efficiency of 100-kVA, single-phase transformer is 98% and occurs at 80% of
full-load at a power factor of 0.8 lagging. If the leakage reactance of the transformer is 5%,
estimate the voltage regulation at rated voltage and a lagging pf of 0.8.
24
A 9.2-kVA, 230/115-V, 50-Hz transformer has the following parameters: ππ = π. πππ +
ππ. πππ π; ππ = π. πππ + ππ. πππ π; βΆ πΉπͺ = πππ. ππ π; πΏ&micro; = πππ. π. For a lagging pf 0.6,
determine: (24.1) Voltage regulation and secondary terminal voltage. (24.2) The full-load
efficiency.
25
A 240/400-V single-phase transformer absorbs 35-W when its LV winding is connected to a
240-V, 50-Hz supply, the secondary being open-circuit. When the LV winding is short-circuited
and a 10-V, 50-Hz supply is connected to HV winding, the power absorbed was 48-W and the
current in the HV winding is 15-A, which is the full-load current. Calculate the: (25.1) Full-load
rating of the transformer (25.2) Efficiency of the transformer when it delivers 40% of rated kVA
at a power factor of 0.8 lagging.
26
A single-phase, 200-kVA, 6351/660-V transformer has the following data: [O.C.T: 6351 Vπ»π
1663π΄π»π, 4.8 πππ»π]; [π. πΆ. π: 500 ππ»π , 30 π΄π»π , 8.2 πππ»π ]. Use the approximate equivalent
diagram and calculate ππΆ , ππ , ππ1 , ππ1 , ππ1 , ΙΈπ .
27
A 20 kVA, 4000/400 V, 50 Hz single-phase transformer has the following parameters:
HV: π = 3 Ω, π = 5 Ω, ππΆ = 20 πΩ, ππ = 1 πΩ.
LV: π = 0.04 Ω π = 0.05 Ω.
Determine: (27.1) The total resistance, leakage reactance
and impedance referred to the HV side (27.2) The total resistance, leakage reactance and
impedance referred to the LV side. (27.3) The full-load copper loss and iron loss of the
transformer (
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