250+ TOP MCQs on OC Test on Transformer and Answers

Transformers Problems on “OC Test on Transformer”.

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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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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:
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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.