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All questions of Three-Phase AC Circuits for Electrical Engineering (EE) Exam
For a polyphase system, the number of Wattmeter required to measure power is equal to _______- a)Number of wires
- b)One less than number of wires
- c)Number of phases
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
For a polyphase system, the number of Wattmeter required to measure power is equal to _______
a)
Number of wires
b)
One less than number of wires
c)
Number of phases
d)
None of the above
|
Rhea Reddy answered |
The number of Wattmeter required to measure power in a polyphase system is determined using Blondell’s theorem. According to this, the number of Wattmeter required is equal to one less than the number of wires in the circuit. For example, in a three phase, four wire system (Star network), the number of Wattmeter required is three.
A polyphase system is generated by______?- a)Having two or more generator windings separated by equal electrical angle.
- b)Having generator windings at equal distances
- c)None of the above
- d)A and C
Correct answer is option 'A'. Can you explain this answer?
A polyphase system is generated by______?
a)
Having two or more generator windings separated by equal electrical angle.
b)
Having generator windings at equal distances
c)
None of the above
d)
A and C
|
|
Rhea Reddy answered |
A generator having two or more electrical windings which are separated by equal electrical angle generates a polyphase electrical system. The electrical angle or displacement depends upon the number of windings or phases. For example, in a three-phase electrical system, the generated voltages are separated from each other by 120° degrees.
For an 8-pole induction motor supplied with power by a 6 pole Alternator at 1200 revolutions per minute, the value of motor speed at slip of 3% is –- a)800 RPM
- b)400 RPM
- c)873 RPM
- d)900 RPM
Correct answer is option 'C'. Can you explain this answer?
For an 8-pole induction motor supplied with power by a 6 pole Alternator at 1200 revolutions per minute, the value of motor speed at slip of 3% is –
a)
800 RPM
b)
400 RPM
c)
873 RPM
d)
900 RPM
|
|
Rhea Reddy answered |
Given
Speed of Alternator, N = 1200 RPM
Number of Alternator poles, P = 6
Supply frequency, f = PN/120 = 60 Hz
Hence, for number of 8 poles in Induction motor, the synchronous speed or speed of rotating field is: Ns = 120f/P = 900 RPM
Let Nr is actual speed of motor.
Thus, Percentage Slip, S % = ((Ns – Nr)/Ns) x 100
Substituting the values, we get: Nr = Ns (1-S/100) = 873 RPM.
In a three phase, delta connection ______.- a)line current is equal to phase current
- b)Line voltage is equal to phase voltage
- c)None of the above
- d)Line voltage and line current is zero
Correct answer is option 'B'. Can you explain this answer?
In a three phase, delta connection ______.
a)
line current is equal to phase current
b)
Line voltage is equal to phase voltage
c)
None of the above
d)
Line voltage and line current is zero
|
|
Rhea Reddy answered |
A delta connected AC circuit is achieved by connecting the start end of a winding to the finish end of another winding such that all three windings form a mesh. Since each end of the windings forms the line connection, voltage across each winding is equal to the potential difference between the corresponding lines taken from that winding. Hence the phase voltage is equal to the line voltage.
For the below star connected network of equal resistances, if the Wattmeter reading is 5kW and ammeter reading is 25 Amperes, the power factor, resistance and inductanceis _______ respectively.
- a)1, 5 Ohms, 0.1H
- b)0.866, 8 Ohms, 0.02H
- c)0.5, 10 Ohms, 0.01H
- d)0.4, 8 Ohms, 0.02H
Correct answer is option 'B'. Can you explain this answer?
For the below star connected network of equal resistances, if the Wattmeter reading is 5kW and ammeter reading is 25 Amperes, the power factor, resistance and inductanceis _______ respectively.
a)
1, 5 Ohms, 0.1H
b)
0.866, 8 Ohms, 0.02H
c)
0.5, 10 Ohms, 0.01H
d)
0.4, 8 Ohms, 0.02H
|
|
Rhea Reddy answered |
Given
Line Voltage, VL = 400 Volts
Frequency, f = 60 Hz
Line Current, IL = 25 Amperes
Power per phase, Pph = 5kW
Phase Voltage, Vph = VL/3^1/2 = 230.9 Volts
Phase current, Iph = 25 Amperes
Hence, power factor, cosφ = Pph/VphIph = 0.866
Impedance, Zph = Vph/Iph = 9.236 Ohms
Resistance, R = Zphcosφ = 8 Ohms
Line Voltage, VL = 400 Volts
Frequency, f = 60 Hz
Line Current, IL = 25 Amperes
Power per phase, Pph = 5kW
Phase Voltage, Vph = VL/3^1/2 = 230.9 Volts
Phase current, Iph = 25 Amperes
Hence, power factor, cosφ = Pph/VphIph = 0.866
Impedance, Zph = Vph/Iph = 9.236 Ohms
Resistance, R = Zphcosφ = 8 Ohms
A polyphase system is generated by ______.- a)Having two or more generator windings separated by equal electrical angle.
- b)Having generator windings at equal distances
- c)None of the above
- d)A and C
Correct answer is option 'A'. Can you explain this answer?
A polyphase system is generated by ______.
a)
Having two or more generator windings separated by equal electrical angle.
b)
Having generator windings at equal distances
c)
None of the above
d)
A and C
|
|
Rhea Reddy answered |
A generator having two or more electrical windings which are separated by equal electrical angle generates a polyphase electrical system. The electrical angle or displacement depends upon the number of windings or phases. For example, in a three-phase electrical system, the generated voltages are separated from each other by 120 degrees.
For a three phase, three wire system, the two Wattmeter read 4000 Watts and 2000 Watts respectively. The power factor when both meters give direct reading is _______ ?- a)1
- b)0.5
- c)0.866
- d)0.6
Correct answer is option 'C'. Can you explain this answer?
For a three phase, three wire system, the two Wattmeter read 4000 Watts and 2000 Watts respectively. The power factor when both meters give direct reading is _______ ?
a)
1
b)
0.5
c)
0.866
d)
0.6
|
|
Rhea Reddy answered |
Reading of Wattmeter 1, W1 = 4000 Watts
Reading of Wattmeter 2, W2 = 2000 Watts
Phase angle;
Power Factor, = 0.866
Power Factor, = 0.866
For a balanced three phase, three wire system with input power of 10kW, at 0.9 power factor, the readings on both wattmeter are _______ respectively.- a)7kW, 3kW
- b)6350W, 3650W
- c)5000W, 5000W
- d)7600W, 1200W
Correct answer is option 'B'. Can you explain this answer?
For a balanced three phase, three wire system with input power of 10kW, at 0.9 power factor, the readings on both wattmeter are _______ respectively.
a)
7kW, 3kW
b)
6350W, 3650W
c)
5000W, 5000W
d)
7600W, 1200W
|
|
Rhea Reddy answered |
Let reading of one Wattmeter = W1
Reading of second Wattmeter = W2
Input Power, P = W1+W2 = √3VLILcosφ = 10000 ________1
Power Factor, cosφ = 0.9
Phase angle, φ = 25.8 degrees
Therefore, W1 = VLILcos(30-φ) = 0.99VLIL = 6350W
W2 = VLILcos(30+φ) = 0.56VLIL = 3650W
Reading of second Wattmeter = W2
Input Power, P = W1+W2 = √3VLILcosφ = 10000 ________1
Power Factor, cosφ = 0.9
Phase angle, φ = 25.8 degrees
Therefore, W1 = VLILcos(30-φ) = 0.99VLIL = 6350W
W2 = VLILcos(30+φ) = 0.56VLIL = 3650W
For a polyphase system, the number of Wattmeter required to measure power is equal to _______.- a)Number of wires
- b)One less than number of wires
- c)Number of phases
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
For a polyphase system, the number of Wattmeter required to measure power is equal to _______.
a)
Number of wires
b)
One less than number of wires
c)
Number of phases
d)
None of the above
|
|
Rhea Reddy answered |
The number of Wattmeter required to measure power in a polyphase system is determined using Blondell’s theorem. According to this, the number of Wattmeter required is equal to one less than the number of wires in the circuit. For example, in a three phase, four wire system (Star network), the number of Wattmeter required is three.
For a three-phase delta connected load, fed from a star connected network, the power transferred to the load is
- a)3 KW
- b)7 KW
- c)5 KW
- d)7KW
Correct answer is option 'B'. Can you explain this answer?
For a three-phase delta connected load, fed from a star connected network, the power transferred to the load is
a)
3 KW
b)
7 KW
c)
5 KW
d)
7KW
|
|
Rhea Reddy answered |
Given values:
Star Connected phase voltage, Vphs = 230 Volts
Phase load resistance, RphLd = 20 Ohms
Phase load reactance, XphLd = 40 Ohms
Star Connected phase voltage, Vphs = 230 Volts
Phase load resistance, RphLd = 20 Ohms
Phase load reactance, XphLd = 40 Ohms
Hence, phase load impedance, = 
Star connected line voltage, VL =√3Vphs = 398.37 Volts
For the delta connected load, Phase Voltage, VphLd = VL = 398.37 Volts
Therefore, current through each phase of load, IphLd = VphLd/ZphLd = 8.9 Amperes
Line current for delta connected load, IL = √3IphLd = 15.41 Amperes
Power Factor, pfs = RphLd/ZphLd = 0.44
Thus, power supplied to load, PL =√3 VLILpfs = 4.7 KW
For the delta connected load, Phase Voltage, VphLd = VL = 398.37 Volts
Therefore, current through each phase of load, IphLd = VphLd/ZphLd = 8.9 Amperes
Line current for delta connected load, IL = √3IphLd = 15.41 Amperes
Power Factor, pfs = RphLd/ZphLd = 0.44
Thus, power supplied to load, PL =√3 VLILpfs = 4.7 KW
Slip of an induction motor lies between ______.- a)0 to 10 %
- b)0 to 400 RPM
- c)0 to 5 %
- d)0 to 4000 RPM
Correct answer is option 'C'. Can you explain this answer?
Slip of an induction motor lies between ______.
a)
0 to 10 %
b)
0 to 400 RPM
c)
0 to 5 %
d)
0 to 4000 RPM
|
|
Rhea Reddy answered |
Slip is the difference between the speed of rotating field and actual speed of rotor. The speed of the rotor must be less than the speed of the rotating field, else there would be no relative motion and the rotary motion would cease to exist. The speed of the rotor should be such that the magnitude of the rotor current is enough to exert the necessary torque. The slip speed denotes the speed of rotor relative to that of the field. Though it is measured in revolutions per minute, more commonly it is denoted in percentage and ranges from 0 to 5%.
An induction motor is ______.- a)Self-starting
- b)Requires external supply
- c)A or C
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
An induction motor is ______.
a)
Self-starting
b)
Requires external supply
c)
A or C
d)
None of these
|
|
Rhea Reddy answered |
As the rotation of the rotor conductors are initiated normally due to the force produced because of interaction between induced current and rotor magnetic field, no external supply is required and the Induction motor is self-starting.
In a three phase, delta connection _________- a)line current is equal to phase current
- b)Line voltage is equal to phase voltage
- c)None of the above
- d)Line voltage and line current is zero
Correct answer is option 'B'. Can you explain this answer?
In a three phase, delta connection _________
a)
line current is equal to phase current
b)
Line voltage is equal to phase voltage
c)
None of the above
d)
Line voltage and line current is zero
|
|
Rhea Reddy answered |
A delta connected AC circuit is achieved by connecting the start end of a winding to the finish end of another winding such that all three windings form a mesh. Since each end of the windings forms the line connection, voltage across each winding is equal to the potential difference between the corresponding lines taken from that winding. Hence the phase voltage is equal to the line voltage.
The name of the Induction motor comes from the fact that ______.- a)Operation of the motor depends upon the induced voltage in the stator
- b)Operation of the motor depends upon the value of coil windings
- c)Operation of the motor depends upon the induced voltage in the rotor conductors
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
The name of the Induction motor comes from the fact that ______.
a)
Operation of the motor depends upon the induced voltage in the stator
b)
Operation of the motor depends upon the value of coil windings
c)
Operation of the motor depends upon the induced voltage in the rotor conductors
d)
None of these
|
|
Rhea Reddy answered |
The rotating magnetic field produced by the stator winding on being subjected to a three-phase supply, causes an induction of voltage in the stationary rotor windings. As the rotor circuit is complete, current starts flowing due to the induced voltage. This induced current in turn produces its own magnetic field. As the current carrying conductors are placed in the magnetic field, a force is produced, which acts tangentially and develops a torque which causes the rotor conductors to rotate. Thus, operation of the motor depends upon the induced voltage in the rotor conductors and the motor is called Induction Motor.
In a squirrel cage induction motor, the rotor slots are- a)Placed in line with the rotor shaft
- b)Skewed at a certain angle to the rotor shaft
- c)Parallel to the rotor shaft
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
In a squirrel cage induction motor, the rotor slots are
a)
Placed in line with the rotor shaft
b)
Skewed at a certain angle to the rotor shaft
c)
Parallel to the rotor shaft
d)
None of these
|
|
Rhea Reddy answered |
A squirrel cage rotor consists of a laminated steel cylinder with slots in which either Aluminium conductors are die-cast of copper bars are used. These bars are shorted at both ends by heavy end rings of same materials. To reduce the magnetic noise and hence produce a more uniform torque and to prevent possible magnetic locking of the rotor with the stator, the rotor slots are skewed at a certain angle to the rotor shaft.
Chapter doubts & questions for Three-Phase AC Circuits - Basic Electrical Technology 2025 is part of Electrical Engineering (EE) exam preparation. The chapters have been prepared according to the Electrical Engineering (EE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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