

MCQOPTIONS
This section includes 403 Mcqs, each offering curated multiple-choice questions to sharpen your Electrical Engineering Questions knowledge and support exam preparation. Choose a topic below to get started.
101. |
While performing sumpner’s test on transformer, if the secondary are not in series opposition then the meter will read |
A. | 0 |
B. | Twice of the induced emf in each secondary’s |
C. | Equal to the induced emf in each secondary’s |
D. | Keep fluctuating |
Answer» C. Equal to the induced emf in each secondary’s | |
102. |
A 10 kVA, 2000 / 100v transformer has R1=1.5 ohm, R2=0.005 ohm, X1=2.5 ohm and X2=0.08 ohm. The equivalent resistance referred to primary is |
A. | 3 ohm |
B. | 3.25 ohm |
C. | 3.5 ohm |
D. | 3.8 ohm |
Answer» D. 3.8 ohm | |
103. |
For maximum efficiency in a transformer |
A. | Core losses = 2 * copper losses |
B. | Core losses = copper losses / 2 |
C. | Core losses = copper losses |
D. | Core losses = copper losses ∧ 2 |
Answer» D. Core losses = copper losses ∧ 2 | |
104. |
A 600 kVA transformer has iron losses of 400 kW and copper losses of 500 kW. Its kVA rating for maximum efficiency is given by |
A. | 537 kVA |
B. | 548 kVA |
C. | 555 kVA |
D. | 585 kVA |
Answer» B. 548 kVA | |
105. |
If the applied voltage of a certain transformer is increased by 50% keeping the frequency constant, the maximum flux density will become |
A. | Remains same |
B. | 1.5 times the original value |
C. | Double the original value |
D. | Half of the original value |
Answer» C. Double the original value | |
106. |
For parallel operation of transformers, the most satisfactory condition is achieved when the load shared by the transformers is in exact proportion with their |
A. | Efficiency |
B. | Ratings |
C. | Impedances |
D. | None of these |
Answer» C. Impedances | |
107. |
If the supply voltage is kept at constant value and the supply frequency is increased. The magnetizing and core loss component |
A. | Both will decrease |
B. | Magnetizing component will increase and core loss will decrease |
C. | Magnetizing component will decrease and core loss will increase |
D. | Both will increase |
Answer» B. Magnetizing component will increase and core loss will decrease | |
108. |
For a certain transformer if E1 is the emf across primary winding, E2 is the emf across secondary winding and K is the ratio of secondary to primary turns. The value of secondary emf referred to the primary side is |
A. | E2 / K |
B. | E2 / K ∧ 2 |
C. | K * E2 |
D. | K ∧ 2 * E2 |
Answer» B. E2 / K ∧ 2 | |
109. |
For a single phase no load transformer, which among the following losses will be minimum? |
A. | hysteresis losses |
B. | eddy current losses |
C. | copper losses |
D. | mechanical losses |
Answer» D. mechanical losses | |
110. |
Which test is also called as back to back test? |
A. | Retardation test |
B. | Sumpner’s test |
C. | Field test |
D. | Voltage drop test |
Answer» C. Field test | |
111. |
The globules of water present in oil may result directly in a breakdown. This can be prevented by |
A. | Proper insulation |
B. | Isolation of oil from air by using an inert gas |
C. | Heating the tank |
D. | None of these |
Answer» C. Heating the tank | |
112. |
In a transformer tank, grounding is given to protect it from lightening stoke. Grounding lug is usually installed near |
A. | Bottom of the tank |
B. | Top of the tank |
C. | In the middle of the tank |
D. | Anywhere on the tank |
Answer» B. Top of the tank | |
113. |
In a step-up transformer when the voltage delivered is higher than that received, the current delivered is |
A. | Higher than that received |
B. | Lower than that received |
C. | Equal to that received |
D. | Equal to half of that received |
Answer» C. Equal to that received | |
114. |
What happens if dc supply is given to the transformer? |
A. | Transformer will work more efficiently |
B. | It will work less efficiently |
C. | It will work same as with ac supply |
D. | Not work at all |
Answer» E. | |
115. |
For a transformer, permeability and hysteresis loss should be |
A. | Low, low |
B. | Low, high |
C. | High, low |
D. | High, high |
Answer» D. High, high | |
116. |
During starting, the magnetic field produced in a transformer is |
A. | 1.5 times the maximum |
B. | Equal to the maximum |
C. | 0.5 times of the maximum |
D. | Equal to zero |
Answer» E. | |
117. |
For transformers ratings are always given in kVA, not in kW because |
A. | It is conventional |
B. | Power factor of load is not known |
C. | For higher efficiency |
D. | To make calculation simple |
Answer» C. For higher efficiency | |
118. |
With increase in load, the speed of synchronous motor |
A. | Increases |
B. | Decreases |
C. | Remains constant |
D. | None of the above |
Answer» D. None of the above | |
119. |
The method used for starting synchronous motor is/are |
A. | Using pony motors |
B. | Using damper windings |
C. | As a slip ring induction motor |
D. | All of the above |
Answer» E. | |
120. |
In case of hydro generators, the driving torque can be changed by controlling the |
A. | Gate opening |
B. | Throttle opening |
C. | Both (a) and (b) |
D. | None of these |
Answer» B. Throttle opening | |
121. |
A synchronous motor is started by using damper winding. When the motor runs at synchronous speed then the induced emf in the damper winding will be |
A. | Zero |
B. | Maximum |
C. | Half of the terminal voltage |
D. | None of these |
Answer» B. Maximum | |
122. |
Back emf in case of synchronous motor depends on the |
A. | Excitation given to the field winding |
B. | Speed |
C. | Both (a) and (b) |
D. | None of the above |
Answer» B. Speed | |
123. |
If the load on a synchronous motor increases then its |
A. | Current drawn by motor increases and load angle decreases |
B. | Current drawn by motor decreases and load angle remains constant |
C. | Current drawn by motor remains constant and load angle increases |
D. | Current drawn by motor increases and load angle also increases |
Answer» E. | |
124. |
According to Blondel’s two reaction theory the armature mmf is divided into components, direct axis component and quadrature axis component. The component acting along quadrature axis is |
A. | Magnetizing |
B. | Demagnetizing |
C. | Cross magnetizing |
D. | None of these |
Answer» D. None of these | |
125. |
When an alternator is connected to infinite bus bar, change in excitation for alternator will |
A. | Change the terminal voltage and power factor both |
B. | Affect only terminal voltage and power factor remains unaffected |
C. | Affect only power factor and terminal voltage remains unaffected |
D. | Neither terminal voltage will get affected nor power factor |
Answer» D. Neither terminal voltage will get affected nor power factor | |
126. |
The increase in mechanical torque in case of alternator will |
A. | Increase the armature current and reduce the power factor |
B. | Increase the armature current and improve the power factor |
C. | Decrease the armature current and improve the power factor |
D. | Decrease the armature current and reduce the power factor |
Answer» C. Decrease the armature current and improve the power factor | |
127. |
If a three phase winding is displaced in space by 120 degree in a synchronous motor then the magnitude of resultant flux when it is rotated by Θ = 0 degree and Θ = 60 degree is |
A. | 0, 1.5 times the magnitude of maximum flux |
B. | 1.5 times the magnitude of maximum flux, 0 |
C. | 0.866 times the magnitude of maximum flux, 1.5 times the magnitude of maximum flux |
D. | 1.5 times the magnitude of maximum flux, 1.5 times the magnitude of maximum flux |
Answer» E. | |
128. |
Hunting in a synchronous motor results in |
A. | Loss of synchronism |
B. | Mechanical and electrical stresses on motor |
C. | Large current drawn by motor |
D. | all of the above |
E. | None of the above |
Answer» E. None of the above | |
129. |
For a half cycle of fluxes in a synchronous motor, the resultant flux has rotated through |
A. | 0 degree |
B. | 90 degree |
C. | 180 degree |
D. | 360 degree |
Answer» D. 360 degree | |
130. |
The reluctance offered to the mmf wave is lowest when |
A. | It is aligned with the field pole axis |
B. | It is oriented at 90 degree to the field pole axis |
C. | Both (a) and (b) |
D. | None of these |
Answer» B. It is oriented at 90 degree to the field pole axis | |
131. |
According to Blondel’s two reaction theory, the armature mmf component acting along direct axis can be |
A. | Demagnetizing |
B. | Magnetizing |
C. | Cross magnetizing |
D. | Either (a) or (b) |
Answer» E. | |
132. |
The synchronous reactance of an alternator is equal to |
A. | Leakage reactance of armature |
B. | Armature reaction reactance |
C. | Sum of leakage reactance and armature reaction reactance |
D. | Difference of leakage reactance and armature reaction reactance |
Answer» D. Difference of leakage reactance and armature reaction reactance | |
133. |
If the frequencies of the two alternators in lamp dark method for synchronization of alternators are unequal, then the two lamps will |
A. | Glow together |
B. | Become alternately bright and dark |
C. | Not glow at all |
D. | None of the above |
Answer» C. Not glow at all | |
134. |
In lamps dark method for synchronization of single phase alternator, if the frequency of two alternators is exactly same as well as their voltages are in exact phase opposition then the resultant voltage will be |
A. | Sum of the voltages of two alternators |
B. | Equal to the voltage of alternators |
C. | Zero |
D. | None of these |
Answer» D. None of these | |
135. |
For synchronization of single phase alternators which method gives the more accurate result? |
A. | Lamps dark method |
B. | Lamps bright method |
C. | Both (a) and (b) |
D. | None of these |
Answer» C. Both (a) and (b) | |
136. |
Which method for finding percentage regulation in synchronous machines is called pessimistic method? |
A. | E.M.F. method |
B. | M.M.F. method |
C. | Z.P.F. method |
D. | Direct loading method |
Answer» B. M.M.F. method | |
137. |
For high capacity alternators, which method is not suitable for finding percentage regulation? |
A. | Direct loading method |
B. | Synchronous impedance method |
C. | Ampere-turns method |
D. | Potier triangle method |
Answer» B. Synchronous impedance method | |
138. |
Low value of SCR in an alternator indicates |
A. | High stability limit |
B. | Low stability limit |
C. | None of these |
Answer» C. None of these | |
139. |
For finding voltage regulation of an alternator, the method which gives most accurate result is |
A. | E.M.F method |
B. | M.M.F. method |
C. | Z.P.F method |
D. | ASA modification of M.M.F. method |
Answer» D. ASA modification of M.M.F. method | |
140. |
If the value of SCR in an alternator is low then the air gap and voltage drop across synchronous reactance will be |
A. | Low, high |
B. | High, low |
C. | Low, low |
D. | High, high |
Answer» B. High, low | |
141. |
In Potier’s triangle method, to determine armature leakage reactance and armature reaction mmf separately, the tests performed are |
A. | Open circuit test and short circuit test |
B. | Open circuit test and zero power factor test |
C. | Short circuit test and zero power factor test |
D. | Open circuit test, short circuit test and zero power factor test |
Answer» C. Short circuit test and zero power factor test | |
142. |
For reducing the slot harmonics in an alternator, the length of air gap is |
A. | Reduced |
B. | Increased |
C. | Not related to harmonics |
D. | None of these |
Answer» C. Not related to harmonics | |
143. |
The open circuit voltage and short circuit current of a 3 phase, star connected alternator are 1050 V and 250 A respectively. Its field current is 12 A. Then the synchronous impedance of alternator will be |
A. | 1.34 ohm |
B. | 1.69 ohm |
C. | 2.42 ohm |
D. | 2.85 ohm |
Answer» D. 2.85 ohm | |
144. |
When the voltage across any two terminals of an armature winding of star connected alternator, then the armature resistance per phase is |
A. | Half of the resistance observed across any two line terminals |
B. | 3/2 of the resistance observed across any two line terminals |
C. | 2/3 of the resistance observed across any two line terminals |
D. | Same as the resistance observed across any two line terminals |
Answer» B. 3/2 of the resistance observed across any two line terminals | |
145. |
If x is the order of harmonic component, then the general expression for distribution factor is |
A. | xsin(β / 2m) / msin(β / 2x) |
B. | sin(x β /m) / xmsin(m β / 2) |
C. | sin(xm β / 2) / msin (x β / 2) |
D. | xmsin(β / 2) /sin(mx β) |
Answer» D. xmsin(β / 2) /sin(mx β) | |
146. |
To minimize the harmonics from the voltage waveform, the suitable steps is / are |
A. | Using distributed type of winding instead of concentrated type |
B. | Using fractional slot windings |
C. | Skewing the pole face |
D. | All of these |
E. | None of these |
Answer» E. None of these | |
147. |
Generally the effective resistance of alternator under a.c. condition is taken as |
A. | Less than 1.25 times the d.c. resistance |
B. | 1.25 to 1.75 times the d.c. resistance |
C. | Greater than 1.75 times the d.c. resistance |
D. | None of these |
Answer» C. Greater than 1.75 times the d.c. resistance | |
148. |
The leakage flux in the armature of alternator makes the armature winding |
A. | Inductive in nature |
B. | Capacitive in nature |
C. | Resistive in nature |
D. | All of these |
Answer» B. Capacitive in nature | |
149. |
Salient pole alternators are best suitable for the application of |
A. | Hydro electric power plant |
B. | Thermal power plant |
C. | Tidal power plant |
D. | Nuclear power plant |
Answer» B. Thermal power plant | |
150. |
The air gap in a salient pole machine is |
A. | Maximum at the center of poles |
B. | Least at the center of poles and increases while moving away from the center |
C. | Maximum at the center and decreases while moving away from the center |
D. | Equally distributed |
Answer» C. Maximum at the center and decreases while moving away from the center | |