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Transformer’s interview questions paper-7

111. What do you mean by resistance drop, reactance drop and impedance drop?

Ans : Resistance drop is the voltage drop in the resistance of the winding which is in phase with the current (IR1’). Reactance drop is the voltage drop in the reactance of the winding which is in quadrature with the current(IX1’). Impedance drop is the vector sum of the resistance drop and the reactance drop.

112.What are the expression of find equivalent circuit ?

Ans: V1 = N1/N2 [I2(r2’+jx2’)+V2]
V2 = N2/N1 [V1 -I1 (r1’ + jx1’)]

113. Which voltage drop becomes larger in a transformer and why?

Ans : The reactance voltage drop becomes larger because of the large value of the inductive reactance of the transformer winding than its ohmic resistance.

114. How are core losses determined in a transformer ?

Ans : Core losses are determined by open circuit test. In this test normal voltage is applied on the low voltage side at rated frequency and since there is no load on the high voltage side the reading of the wattmeter connected on the low voltage side gives the no load loss or core losses of the transformer. High voltage side is kept open due to suitability of using metering arrangement in the low voltage side rather than high voltage side.

115. Why are copper losses neglected open circuit test?

Ans : Since no load current is very small and no load copper loss is very small in comparison with iron losses so the copper losses are neglected in open circuit test.

116. How are the copper losses determined in a transformer ?

Ans : Copper losses can be determined by short circuit test. In this test the low voltage secondary of the transformer is short circuited through an ammeter. At first a small voltage is applied to the high voltage primary side and secondary. Since the test voltage is very small in comparison to the normal voltage the core losses are negligible and the wattmeter connected on the primary side records the full load copper losses.

117. How is the copper loss affected by power factor ?

Ans : Copper loss varies inversely with the power factor.

118. In a short circuit test if the wattmeter reads 160 watts at 20 amperes of current what will be the reading of the wattmeter at 50 amperes of current ?

Ans : W =160 × (50/20 )2=1000 w

119. Why is a C.T used with the wattmeter when performing a short circuit test on a transformer ?

Ans : Since the current coil of a wattmeter is capable of carrying about 5 amps therefore during short circuit test the current more than 5 amps will cause damage to the current coil. So for safety of the current coil of the wattmeter a current transformer is used.

120. What is impedance voltage ?

Ans: The impedance voltage of a transformer required to circulate rated current through a winding of the transformer when another winding is short circuited with the respective windings connected as for rated voltage operation. It is usually expressed in percent of the rated voltage of the winding in which the voltage is measured.

121. What do you understand by the efficiency of a transformer ?

Ans: The ordinary or commercial efficiency of a transformer is defined as the ratio of the output in watts to the input in watts. Since the efficiency is based on the power output in watts and not on volt-amperes, the power factor plays an important role in determining the efficiency.
Efficiency =(Output power)/(Input power) =(E2 I2×P.F)/(E2 I2×P.F+Losses)
= (E2 I2 cos∅)/(E2 I2 cos∅+W1+I1’ R1+I2’ R2 )
Where E2 =Secondary voltage.
I2 = Secondary current.
I1 =Primary current.
R1 =Primary resistance.
R2 = Secondary resistance.
W1 =Iron losses.

122. What is the condition for the maximum efficiency of a transformer ?

Ans : The condition for the maximum efficiency of a transformer is (Iron losses =copper losses).

123. On what factors does the efficiency of a power transformer depend ?

Ans : The efficiency of a power transformer depends on the power factor of the load and the percentage of the load.

124. How does the efficiency of a power transformer depend on the load current ?

Ans : The efficiency increases with the increase in load current and reaches to maximum value and then decreases with further increase in load current .

125. A transformer has iron loss of 300 watts and full load copper loss of 500 watts. What will be the total loss of the transformer for maximum efficiency ?

Ans : 600 watts.

126. Under what condition of losses in a transformer maximum efficiency be obtained at 7/8 full load ?

Ans :When copper loss =64/49 × Iron losses.

127. Why is the efficiency of a transformer so high ?

Ans : Since the transformer is a static apparatus without any rotating parts there frictional loss. The only losses are iron losses end copper losses in the primary and secondary. Therefore, the efficiency of a transformer becomes so high about 95% to 98%.

128. What is the all day efficiency of a transformer ?

Ans : All day efficiency is defined as the ratio of energy output in 24 hours to energy input in 24 hours.

129. What is meant by regulation of a transformer ?

Ans : The regulation of a transformer is the difference between the no load and full load secondary voltages expressed in terms of the no load voltage with constant primary voltage.

130. How is the percentage regulation of a transformer is calculated ?


When, R= Equivalent resistance in terms of secondary.
X =Equivalent resistance in terms of secondary.
E = Induced e.m.f. of secondary.
Φ = Phase difference between voltage and current.
I =Full load current.

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