Up To Date Transformers MCQs – Transformer Losses & Testing ( Transformers ) MCQs
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Transformers MCQs – Transformer Losses & Testing ( Transformers ) MCQs
The most occurred mcqs of Transformer Losses & Testing ( ) in past papers. Past papers of Transformer Losses & Testing ( Transformers ) Mcqs. Past papers of Transformer Losses & Testing ( Transformers ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Transformer Losses & Testing ( Transformers ) Mcqs. The Important series of Transformer Losses & Testing ( Transformers ) Mcqs are given below:
Transformer Losses
1. In a given transformer for given applied voltage, which of the following losses remain constant irrespective of load changes?
a) Friction and windage losses
b) Copper losses
c) Hysteresis and eddy current losses
d) Cannot be determined
Answer: c
Explanation: Hysteresis and eddy current losses together called as core-loss in a transformer. These losses remain constant for constant voltage and frequency applied to a transformer, these components remain same irrespective of load.
2. On which of the following degree of mechanical vibrations produced by the laminations of a transformer depends?
a) Tightness of clamping
b) Gauge of laminations
c) Size of laminations
d) Tightness of clamping, gauge and size of laminations
Answer: d
Explanation: Mechanical vibrations produced in a transformer are directly effective due to the tightness of the clamping, gauge og laminations, size of laminations as well. There are various methods in order to reduce their effects.
3. Variations in a hysteresis loss in a transformer (Bmax = maximum flux density) ____________
a) Bmax
b) Bmax1.6
c) Bmax3.83
d) Bmax/2
Answer: b
Explanation: According to Steinmetz’s formula, the heat energy dissipated due to hysteresis is given by Wh=ηβmax1.6, and hysteresis loss is thus given by Ph≈ Whf ≈ηfβmax1.6. That exponetital term varies fraom 1.4 -1.8 and is equal to 1.6 for iron.
4. Leakage flux in the transformer depends on _____________________
a) Load current
b) Load current and voltage
c) Load current, voltage and frequency
d) Load current, voltage, frequency and power factor
Answer: a
Explanation: Leakage flux is directly proportional to the current, as if is current increased net value of flux increases thus, flux leakage also increases which further contribute to the losses as it is then not able to link with secondary windings.
5. The full-load copper loss of a transformer is 1600 W. At half-load, the copper loss will be _______
a) 6400 W
b) 1600 W
c) 800 W
d) 400 W
Answer: d
Explanation: Copper losses are defined as I2*R losses many times, as they are directly proportional to the square of current flowing through them. Thus, copper losses will reduce if load is reduced that too in square proportion.
6. Silicon steel used in laminations, because it reduces ________________
a) Hysteresis loss
b) Eddy current losses
c) Copper losses
d) Cannot be determined
Answer: a
Explanation: Electrical steels are also known as lamination steel or silicon steel. The main special thing related to the silicon steel is, its magnetic properties such as small hysteresis area and hence, small energy dissipation per cycle, thus low core loss.
7. If the supply frequency to the transformer is increased, the iron loss will ___________
a) Not change
b) Decrease
c) Increase
d) Cannot be determined
Answer: c
Explanation: As frequency increases, the flux density in the core decreases but as the iron loss is directly proportional to the frequency hence effect of increased frequency will be reflected in increase of the iron losses.
8. Which of the following can measure iron loss of a transformer?
a) Low power factor wattmeter
b) Unity power factor wattmeter
c) Frequency meter
d) Any type of wattmeter
Answer: a
Explanation: As the secondary side is open in OC, the entire coil will be purely inductive in nature. So, the power will be lagging due to inductive property of the circuit. So LPF (Low power factor) wattmeter is used in open circuit test of transformer.
9. How reduction in core losses and increase in permeability can be obtained simultaneously in a transformer?
a) Core built-up of laminations of cold rolled grain oriented steel
b) Core built-up of laminations of hot rolled sheet
c) Cannot be determined
d) Frequency Meter
Answer: a
Explanation: CRGO is supplied by the producing mills in coil form and has to be cut into laminations, which are then used in transformer core, which is an integral part of any transformer. Grain-oriented steel is used in large power, distribution transformers and in certain audio output transformers also.
10. Losses which occur in rotating electric machines and do not occur in transformer are ______
a) Friction and windage losses
b) Magnetic losses
c) Hysteresis and eddy current losses
d) Copper losses
Answer: a
Explanation: Windage and friction losses occur in rotating parts of a machine generally in rotor of the machine, thus they will never occur in transformer, as transformer does not contain any rotating part at its secondary unlike induction motor.
Real Transformer And Equivalent Circuit MCQs
11. In a given transformer for a given applied voltage, which losses remain constant irrespective of load changes?
a) Hysteresis and eddy current losses
b) Friction and windage losses
c) Copper losses
d) Cannot be determined
Answer: a
Explanation: Hysteresis and eddy current losses are voltage and frequency dependent losses that too from primary side thus, load change will not make any effect on these losses and they will remain constant as long as voltage and frequency is constant.
12. Which of the following loss in a transformer is zero even at full load?
a) Core loss
b) Friction loss
c) Eddy current loss
d) Hysteresis loss
Answer: b
Explanation: Friction losses are involved with rotating parts of a machine. Since in a transformer all parts are stationary, friction losses will always be equal to zero, irrespective of the loading condition.
13. A shell-type transformer has __________
a) High eddy current losses
b) Reduced magnetic leakage
c) Negligible hysteresis losses
d) Cannot be determined
Answer: b
Explanation: Since windings are brought closer in shell type compare to core type transformer, leakage of flux is very less in shell type transformer. Most of the flux gets linked with both of the coils though there is some leakage which can’t be avoided.
OC Test on Transformer
1. During open circuit test (OC) of a transformer _____________
a) primary is supplied rated kVA
b) primary is supplied full-load current
c) primary is supplied current at reduced voltage
d) primary is supplied rated voltage
Answer: d
Explanation: Open circuit test is normally conducted on rated voltage because any machine is constructed to give maximum efficiency near rated value. Hence, it is operated at rated voltage, and we have to perform the test on machine is to be used.
2. Open circuit test on transformers is conducted so as to get ______________
a) Hysteresis losses
b) Copper losses
c) Core losses
d) Eddy current losses
Answer: c
Explanation: Open circuit test gives the core losses also called as iron losses and shunt parameters of the equivalent circuit of transformer. Open circuit test and short circuit test both provide all the parameters of equivalent circuit.
3. Why OC test is performed on LV side?
a) Simple construction
b) Less voltage is required and parameters can be transformed to HV side
c) It’ll not give losses ig conducted on HV side
d) HV side does not have connections for voltage
Answer: b
Explanation: Open circuit test can be performed on any side but for our convenience and supply voltage available we generally conduct the test on LV side, to get corresponding parameters on HV side we can use transformation ratio.
4. In OC test all the power supplied is utilised for ______
a) Core losses
b) Iron losses
c) Windage losses
d) Cannot be determined
Answer: b
Explanation: In open circuit test all the power supplied is used to overcome iron losses and hence, by taking the reading of input power one can easily do the calculations to find shunt parameters of equivalent circuit of transformer.
5. How shunt branch component Gi is calculated?
a) Po/v12
b) V1/Io
c) Io/ V2
d) Any of the above
Answer: c
Explanation: Shunt branch resistance inverse is denoted by Gi. This Gi can be calculated by the power drop taking place in the resistance divided by square of the voltage applied across the resistor. Current by voltage will give net admittance.
6. Which of the following statements is/are correct statements?
a) EMF per turn in LV winding is more than EMF per turn in LV winding
b) EMF per turn in LV winding is less than EMF per turn in LV winding
c) EMF per turn in HV winding is equal to EMF per turn in LV winding
d) Can’t comment
Answer: c
Explanation: In a transformer, primary volt-ampere is equal to secondary volt-ampere and primary ampere turns are also equal to secondary ampere turns So, EMF per turn in both the winding are equal. Total induced emf on both sides depends on the number of turns, flux and frequency.
7. If the applied voltage of a transformer is increased by 50% and the frequency is reduced by 50%, the maximum flux density will _____________
a) Changes to three times the original value
b) Changes to 1.5 times the original value
c) Changes to 0.5 times the original value
d) Remains the same as the original value
Answer: a
Explanation: Magnetic flux density α β/A. Magnetic flux φ α V/f. φ2/ φ1 = V2/V1 * f1/f2. Since voltage is increased by 50%, V2 thus becomes 1.5 times V1 and frequency becomes 0.5 times the original frequency. Thus, maximum flux density changes to 3 times the original value.
8. The total core loss can be termed as ____________
a) Eddy current loss
b) Hysteresis loss
c) Copper loss
d) Magnetic loss
Answer: d
Explanation: The total core loss is due to iron core or any core material used. As iron loss is proportional to magnetic flux density or flux, these are also called as magnetic loss. The total core loss or magnetic loss in any given transformer totally consists of eddy current loss and hysteresis loss.
9. 2 KVA, 230 V, 50 Hz single phase transformer has an eddy current loss of 40 watts. The eddy current loss when the transformer is excited by a dc source of same voltage will be ___________
a) Equal to 40W
b) Less than 40W
c) More than 40W
d) Zero watts
Answer: d
Explanation: Eddy current loss is directly proportional to the frequency^2. So, for DC current frequency is equal to 0 Hz. Thus, eddy current losses being directly proportional to square of frequency they’ll be equal to 0.
10. Which of the following is the correct formula for Bm?
a) Bm= (Yo2-Gi2)(1/2)
b) Bm= (Yo2+Gi2)(1/2)
c) Bm= (Yo2-Gi2)(2)
d) Bm= (Yo2+Gi2)(1/2)
Answer: a
Explanation: We get the value of Y0 from the no-load current and voltage reading as, Io/V1. Similalry we get the value of Gi from output power and voltage reading as, Po/V1. It then follows that, Bm= (Yo2-Gi2)(0.5).
11. How shunt branch component Y0 is calculated?
a) I0/V1
b) V1/I0
c) P0/V12
d) Cannot be determined
Answer: a
Explanation: Shunt branch admittance is defined as inverse of shunt branch impedance. As we know, impedance can be calculated by the simple ohm’s law; admittance is equal to the inverse of the impedance.
SC Test on Transformer
1. While conducting short-circuit test on a transformer which side is short circuited?
a) High voltage side
b) Low voltage side
c) Primary side
d) Secondary side
Answer: b
Explanation: It’s a common practice to conduct SC test from HV side, while keeping LV side short circuited. Thus, short circuited current is made to flow from shorted low voltage terminals i.e. LV side.
2. During short circuit test why iron losses are negligible?
a) The current on secondary side is negligible
b) The voltage on secondary side does not vary
c) The voltage applied on primary side is low
d) Full-load current is not supplied to the transformer
Answer: c
Explanation: Very small amount of voltage is given to the transformer primary thus the magnetic losses which are dependent on magnetic flux density will get minimum value, hence iron losses are negligible.
3. Short circuit test on transformers is conducted to determine ______
a) Core losses
b) Copper losses
c) Hysteresis losses
d) Eddy current losses
Answer: b
Explanation: Short circuit test is used to determine the copper losses taking place in the transformer under operation, while open circuit test gives us the value of core losses taking place in the transformer.
4. When a short circuit test on a transformer is performed at 25 V, 50 Hz, the drawn current is I1. If the test is performed by 25 V and 25 Hz and power drawn current is I2, then
a) I1 > I2
b) I1 < I2
c) I1 = I2
d) Can’t be defined
Answer: b
Explanation: Current by ohm’s law is equal to voltage divided by impedance. So, I=V/Z. Here Z is inductive load, thus Z= 2πfL. So as the frequency decreases the impedance also decreases and ultimately it reduces the denominator term causing increase in current.
5. Why SC test is not conducted on LV side?
a) Difficult to arrange low voltage supply
b) Difficult to arrange high current supply
c) Difficult to arrange low voltage and high current supply to the LV
d) SC test on LV does not give correct results
Answer: c
Explanation: If rated voltages and power is considered we need only 5% of rated voltage to be applied at on HV side, while by calculations current requirement is also less. For the same test on LV side though voltage required is less compare to HV side, current required is very high.
6. SC test gives ______________
a) Series parameters of equivalent circuit
b) Parallel parameters of equivalent circuit
c) Both parameters of equivalent circuit
d) Neither series nor parallel parameter of equivalent circuit
Answer: a
Explanation: Short circuit test gives the copper losses; these losses are taken into consideration by series parameters of the equivalent circuit. While, Open circuit test gives us iron losses; which are shown by parallel components of equivalent circuit.
7. For 200 kVA, 440/6600-V transformer, short circuit test on the LV side would require ______
a) 22V
b) 330V
c) 44V
d) Can’t be calculated
Answer: a
Explanation: For a given transformer SC test is conducted on LV side, thus we’ll use 5% of rated voltage on the low voltage side. Hence, 5% of 440V calculation gives the value of 440*5/100= 22V on LV side.
8. For a transformer given of 100 kVA, 220/6000-V transformer, short circuit test is performed. What current rating is needed?
a) 30A
b) 445A
c) 60A
d) Can’t be calculated
Answer: b
Explanation: For a given transformer here, test is performed on low voltage side, thus we need the value of current on the low voltage side, by dividing the reactive power by the rated voltage value, i.e. 200*1000/100= 445A.
9. What will be the value of voltage and current for a given transformer of 10 MVA, 220/4400-V which we are about to perform the Short circuit test?
a) 220 V, 30 A
b) 220 V, 2.27A
c) 440 V, 30 A
d) 440 V, 2.27 A
Answer: b
Explanation: Since short circuit test is always done on the HV side unless mentioned specifically, thus values are calculated with HV side parameters. Voltage required on HV side = 4400*5/100 = 220 V and 10*1000/4400= 2.27A.
10. We only get copper losses from the short circuit test.
a) True
b) False
Answer: a
Explanation: Since the transformer is excited at very low voltage, the iron-loss is negligible (that is why shunt branch is left out), the power input corresponds only to the copper-loss, i.e. PSC = PC (copper-loss).
11. With the help of short circuit calculations we get value of ____________
a) Individual resistance and inductance of both sides
b) Resistance and inductance of primary side
c) Resistance and inductance of primary side
d) Combined resistance and inductance of both sides
Answer: d
Explanation: Short calculations include the ratio of short circuited voltage to the short-circuited current which gives Z value, similarly the R value is calculated by dividing the Short-circuited power with short circuited current square. Then, X is calculated for whole circuit.
12. Short circuit test is performed on a transformer with a certain impressed voltage at rated frequency. What will happen if the short circuit test is now performed with the same magnitude of impressed voltage, but at frequency higher than the rated frequency?
a) The magnitude of current will increase, but power factor will decrease
b) The magnitude of current will decrease, but power factor will increase
c) The magnitude of current will increase, power factor will increase
d) The magnitude of current will decrease, power factor will decrease
Answer: d
Explanation: Since frequency has been increased, the leakage reactance will increase. Which will increase the impedance. Thus, current will be less due to inverse proportionality and power factor will be poorer.
Sumpner’s Test
1. Sumpner’s test is conducted on transformers to study effect of ____________
a) Temperature
b) Stray losses
c) All-day efficiency
d) Cannot be determined
Answer: a
Explanation: Sumpner’s test is the test which is used to determine the steady temperature rise if the transformer was fully loaded continuously; this is so because under each of these tests the power loss to which the transformer is subjected is either the core-loss or copper-loss but not both.
2. Which of the following tests are enough to find all the parameters related to a transformer?
a) OC test
b) OC, SC test
c) OC, SC, Sumpner’s test
d) Sumpner’s test
Answer: c
Explanation: While OC and SC tests on a transformer yield its equivalent circuit parameters, these cannot be used for the ‘heat run’ test wherein the purpose is to determine the steady temperature rise if the transformer was fully loaded continuously.
3. Sumpner’s test is performed on _________
a) Single transformer at a time
b) Only two transformers at a time
c) Minimum 2 transformers at a time
d) Many transformers at a time
Answer: b
Explanation: Sumpner’s test is used to determine the effect of transformer on loaded condition. Thus, two transformers are tested simultaneously, where one simply acts as a load to another transformer.
4. In Sumpner’s test __________
a) Primaries can be connected in either way
b) Primaries are connected in parallel with each other
c) Primaries of both transformers are connected in series with each other
d) No need to connect primaries
Answer: b
Explanation: Sumpner’s test is also called as back-to-back test, where two transformers are used where one transformer acts as a load to another transformer. Primaries of both of the transformers used in a test, are connected in parallel with each other.
5. Which test is sufficient for efficiency of two identical transformers under load conditions?
a) Short-circuit test
b) Back-to-back test
c) Open circuit test
d) Any of the above
Answer: b
Explanation: Open circuit test and short circuit test collectively gives the value of all parameters of an equivalent circuit of a transformer. While Sumpner’s back-to-back test gives the heat run effect of machine by considering rise in temperature.
6. In Sumpner’s test _____________
a) Two secondaries are connected in phase opposition
b) Two secondaries are connected in phase addition
c) Can be connected in either way
d) Never connected with each other
Answer: a
Explanation: In conducting Sumpner’s test two primaries are connected in parallel to the rated voltage supply and secondaries are connected in phase opposition. For the secondaries to be in phase opposition rated secondary voltage across the terminals to be zero.
7. When secondaries are connected in phase opposition, power drawn by the circuit is equal to ___________
a) 2*Pi
b) Pi2
c) Pi
d) 2*Pc
Answer: a
Explanation: If V2 source is assumed shorted, the two transformers appear in open circuit to source V1 as their secondaries are in phase opposition and therefore no current can flow in them. The current drawn from source V1 is thus 2I0 and power is 2P0 = 2Pi, twice the core-loss of each transformer.
8.When the AC supply at the primary side of a transformer are shorted, power drawn by the circuit is equal to ___________
a) 2*PC
b) 2*Pi
c) 2*PC + 2*Pi
d) Can’t be determined
Answer: a
Explanation: When the ac supply (10) terminals are shorted, the transformers are series-connected across V2 supply and are short-circuited on the side of primaries. Therefore, the impedance seen at V2 is 2Z and when V2 is adjusted to circulate full-load current (Ifl), the power fed in is 2PC (twice the full-load copper-loss of each transformer).
9.Total power required for Sumpner’s test is given by ________
a) PC + Pi
b) PC + 2Pi
c) 2PC + Pi
d) 2(PC + Pi)
Answer: d
Explanation: In the Sumpner’s test while the transformers are not supplying any load, full iron-loss occurs in their cores and full copper-loss occurs in their windings; net power input to the transformers being (2P0 + 2Pc). The heat run test could, therefore, be conducted on the two transformers, while only losses are supplied.