This set of Transformers Interview Questions and Answers for Experienced people focuses on “Load Division Between Transformers in Parallel”.

1. If the primary voltages at two transformers V1 and V2 are not equal then on load, we’ll get ____________

a) V_{1}-V_{2} at secondary

b) E_{1}-E_{2} at secondary

c) V_{1}+V_{2} at secondary

d) E_{1}+E_{2} at secondary

View Answer

Explanation: When two transformers paralleled on both sides with proper polarities but on no-load. The primary voltages V

_{1}and V

_{2}are obviously equal. If the voltage-ratio of the two transformers are not identical, the secondary induced emfs, E1

_{1}and E

_{2}though in phase will not be equal in magnitude and the difference (E

_{1}-E

_{2}) will appear across the switch S.

2. If two transformers’ secondaries are connected to each other with unequal primary voltage ratio then, __________

a) no circulating current will flow

b) very high short circuit current will flow

c) small circulating current will flow

d) insufficient information

View Answer

Explanation: When secondaries are paralleled by closing the switch, a circulating current appears even though the secondaries are not supplying any load, as a result of difference in their voltage ratios.

3. The circulating current flowing through the circuit at no load condition depends on ________________

a) total leakage impedance of the two transformers

b) difference in their voltage ratios

c) difference in voltage ratios, leakage impedance of 2 transformers

d) other parameters

View Answer

Explanation: The circulating current flowing at no-load condition depend upon the total leakage impedance of the two transformers and the difference in their voltage ratios. Only a small difference in the voltage-ratios can be tolerated.

4. If the transformers have equal voltage ratio then, ____________

a) exciting current can be neglected

b) summation of two transformer currents is not equal to the net load current

c) difference of two transformer currents is equal to the net load current

d) current will not flow

View Answer

Explanation: When the transformers have equal voltage ratio, E1 = E2, the equivalent circuit of the two transformers would then be simple because of the assumption that the exciting current can be neglected in comparison to the load current.

5. Which is the correct formula for current flowing through the transformer 1, when they’re having equal voltage ratio?

a) I_{1}= Z_{2}/(Z_{1}+Z_{2}) *I_{L}

b) I_{2}= Z_{2}/(Z_{1}+Z_{2}) *I_{L}

c) I_{1}= Z_{1}/(Z_{1}+Z_{2}) *I_{L}

d) I_{1}= Z_{2}/(Z_{1}-Z_{2}) *I_{L}

View Answer

Explanation: Since both the transformers are having same number of turns and thus equal voltage ratios, it is easier for analysis of current as division in two branches will be according to the Ohm’s law, which is the answer.

6. Individual currents are in two loaded transformers ______________

a) inversely proportional to the respective leakage impedances

b) inversely proportional to the net leakage impedances

c) inversely proportional to another leakage impedance

d) directly proportional to the respective leakage impedances

View Answer

Explanation: the individual currents are inversely proportional to the respective leakage impedances. Thus, if the transformers are to divide the total load in proportion to their kVA ratings, it is necessary that the leakage impedances be inversely proportional to the respective kVA ratings.

7. Which of the following is the correct ratio, for transformers having equal voltage ratios?

a) Z_{1}/Z_{2}= S_{2}(rated)/ S_{1}(rated)

b) Z_{1}/Z_{2}= S_{1}(rated)/ S_{2}(rated)

c) Z_{1}/Z_{2}= I_{1}(rated)/ I_{2}(rated)

d) Depends upon the type of connection

View Answer

Explanation: If the transformers are to divide the total load in proportion to their kVA ratings, it is necessary that the leakage impedances be inversely proportional to the respective kVA ratings. Thus, Z

_{1}/Z

_{2}= S

_{2}(rated)/ S

_{1}(rated) = V

_{L}I

_{2}/ V

_{L}I

_{1}. Hence, Z

_{1}/Z

_{2}= I

_{2}/ I

_{1}.

8. Which of the following is the correct statement?

a) S_{L}(max)= S_{2}(rated) Z_{1}+Z_{2}/Z_{2}

b) S_{L}(max)= S1(rated) Z_{1}+Z_{2}/Z_{1}

c) S_{L}(max)= S_{2}(rated) Z_{1}+Z_{2}/Z_{1}

d) S_{L}(max)= S_{1}(rated)+S_{2}(rated)

View Answer

Explanation: We can define maximum load in kVA relating with rated kVA of transformer 1 as, S

_{2}(rated)= (Z

_{1}/Z

_{2}+Z1) *S

_{L}(max). Thus, by rearranging terms we get S

_{L}(max)= S

_{2}(rated) Z

_{1}+Z

_{2}/Z

_{1}and SL(max)= S

_{1}(rated) Z1+Z

_{2}/Z

_{2}.

9. Maximum load kVA is _____________________

a) greater than addition of individual rated kVAs

b) lesser than addition of individual rated kVAs

c) equal to addition of individual rated kVAs

d) depends on the loading condition

View Answer

Explanation: As S

_{L}(max)= S2(rated) Z

_{1}+Z

_{2}/Z

_{1}and S

_{L}(max)= S

_{1}(rated) Z

_{1}+Z

_{2}/Z

_{2}, because of individual leakage impedances are inversely proportional to the respective kVA ratings. Important thing is in either of the above cases maximum kVA loading is lesser than the addition of both rated kVAs.

10. Which is the correct formula of current flowing through one of the transformers having unequal ratios?

a) I_{2}= E_{2}Z_{1}-(E_{1}-E_{2}) Z_{L} / (Z_{1}Z_{2} + ZL(Z_{1}+Z_{2}))

b) I_{1}= E_{2}Z_{1}+(E_{1}+E_{2}) Z_{L} / (Z_{1}Z_{2} + ZL(Z_{1}+Z_{2}))

c) I_{1}= E_{2}Z_{1}-(E_{1}-E_{2}) Z_{L} / (Z_{1}Z_{2} – ZL(Z_{1}-Z_{2}))

d) I_{2}= E_{2}Z_{1}+(E_{1}-E_{2}) Z_{L} / (Z_{1}Z_{2} + ZL(Z_{1}+Z_{2}))

View Answer

Explanation: We know that a small difference in voltage ratios can be tolerated in the parallel operation of transformers. Thus, in the unequal voltage ratio condition current will flow from depending on both of the options stated above.

11. A 600-kVA, single-phase transformer with 0.012 pu resistance and 0.06 pu reactance is connected in parallel with a 300-kVA transformer with 0.014 pu resistance and 0.045 pu reactance to share a load of 800 kVA at 0.8 pf lagging. Find how they share the load (a) when both the secondary voltages are 440 V.

a) S_{1}= 377+j305.2

b) S_{2}= 377-j305.2

c) S_{1}= 264+ j171.6

d) S_{1}= 377-j305.2

View Answer

Explanation: Z

_{1}= 0.012+j 0.06

Z

_{2}= 2(0.014+j0.045)

Z

_{1}+Z2= 0.04+ j0.15

The load is SL = 800(0.8-0.6j)

Thus, S

_{1}= Z

_{2}/Z

_{1}+Z

_{2}S

_{L}, we get S

_{1}= 377- j305.2

S

_{2}= Z

_{1}/Z

_{1}+Z

_{2}S

_{L}, we get S

_{2}= 264-j171.6.

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