71. Calculate the total heat dissipated in a rotor resistor of 14.23 Ω when.65 A current flows through it.
A. 6.45 W
B. 6.01 W
C. 6.78 W
D. 6.98 W
Answer: B
The rotor resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in the case of the resistor so the angle between V & I is 90°.
P=I2R=.65×.65×14.23=6.01 W.
72. Calculate mark to space ratio if the system is on for 4.3 sec and off for 78.2 sec.
A. .054
B. .047
C. .039
D. .018
Answer: A
Mark to space is Ton÷Toff. It is the ratio of the time for which the system is active and the time for which is inactive.
M = Ton÷Toff = 4.3÷78.2 = .054.
73. In the rotor voltage injection method, when an external voltage source is in phase with the main voltage then speed will ___________
A. Increase
B. Decrease
C. Remain unchanged
D. First increases then decrease
Answer: A
In the rotor injection method, when an external voltage is in phase with the main voltage net voltage increases and the value of slip decreases, and the value of rotor speed increases.
74. A 2-pole, 3-phase, ______ Hz induction motor is operating at a speed of 550 rpm. The frequency of the rotor current of the motor in Hz is 2.
A. 9.98
B. 9.71
C. 9.12
D. 9.37
Answer: D
Given a number of poles = 2.
The rotor speed is 550 rpm.
Ns=120×f÷P=120×f÷2 = 60f rpm.
S=Ns-Nr÷Ns. F2=sf. S=F2÷f.
The supply frequency is 9.37 Hz.
75. Calculate the average value of the sinusoidal waveform x(t)=848sin(1.65πt+2π÷0.68).
A. 0
B. 78 V
C. 15 V
D. 85 V
Answer: A
The sinusoidal waveform is generally expressed in the form of
V=Vmsin(Ωt+α)
where
Vm represents peak value
Ω represents angular frequency
α represents a phase difference.
The average value of a sine wave is zero because of equal and opposite lobes areas.
76. R.M.S value of the periodic square waveform of amplitude 72 V.
A. 72 V
B. 56 V
C. 12 V
D. 33 V
Answer: A
R.M.S value of the periodic square waveform is Vm and r.m.s value of the trapezoidal waveform is Vm÷3½.
The peak value of the periodic square waveform is Vm.
77. In the rotor voltage injection method, when an external voltage source is in the opposite phase to the main voltage then speed will ___________
A. Increase
B. Decrease
C. Remain unchanged
D. First increases then decrease
Answer: B
In the rotor injection method, when an external voltage is in the opposite phase to the main voltage net voltage decreases and the value of slip increases, and the value of rotor speed decreases.
78. The rotor injection method is a part of the slip-changing technique.
A. True
B. False
Answer: A
The rotor injection method comes under the slip changing technique. It uses an external voltage source to change the slip value. The load torque remains constant here.
79. The slip recovery scheme is a part of the synchronous speed-changing technique.
A. True
B. False
Answer: A
The slip recovery scheme comes under the synchronous speed changing technique. It uses an external induction machine to change the frequency value. The load torque remains constant here.
80. The slope of the V-I curve is 9.1°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
A. .16 Ω
B. .26 Ω
C. .25 Ω
D. .44 Ω
Answer: A
The slope of the V-I curve is resistance.
The slope given is 9.1° so R=tan(9.1°)=.16 Ω.
The slope of the I-V curve is reciprocal to resistance.
81. If induction motor air gap power is 1.8 KW and gross developed power is .1 KW, then rotor ohmic loss will be _________ KW.
A. 1.7
B. 2.7
C. 3.7
D. 4.7
Answer: A
Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power.
Rotor ohmic losses=Air gap power-Mechanical developed power = 1.8-.1=1.7 KW.
82. The power factor of a squirrel cage induction motor is ___________
A. High at light load only
B. High at heavy loads only
C. Low at the light and heavy loads both
D. Low at rate load only
Answer: B
At heavy loads, the current drawn is high due to which the active power component increases. An increase in the active power component increases the power factor of the machine.
83. At low values of slip, the electromagnetic torque is directly proportional to ___________
A. s
B. s2
C. s3
D. s4
Answer: A
At low values of slip, the electromagnetic torque is directly proportional to the slip value. Due to heavy loading slip value decreases which increases the ratio of R2÷s.
84. Calculate the time period of the waveform v(t)=12sin(8πt+8π÷15)+144sin(2πt+π÷6)+ 445sin(πt+7π÷6).
A. 8 sec
B. 4 sec
C. 7 sec
D. 3 sec
Answer: B
The fundamental time period of the sine wave is 2π.
The time period of z(t) is L.C.M {4,1,2}=4 sec.
The time period is independent of phase shifting and time-shifting.
85. Calculate the total heat dissipated in a resistor of 44 Ω when 0 A current flows through it.
A. 0 W
B. 2 W
C. 1.5 W
D. .3 W
Answer: A
The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in the case of the resistor so the angle between V & I is 90°.
P=I2R=0×0×44=0 W.
86. The value of slip at which maximum torque occurs ________
A. R2÷X2
B. 4R2÷X2
C. 2R2÷X2
D. R2÷3X2
Answer: A
The maximum torque occurs when the slip value is equal to R2÷X2. Maximum torque is also known as breakdown torque, stalling torque, and pull-out torque.
87. Calculate the voltage regulation in the synchronous machine if the no-load voltage is 12 V and the full load voltage is 15V.
A. -20%
B. -40%
C. -60%
D. -80%
Answer: A
Voltage regulation is defined as the fluctuation in the load voltage when the load is varied from no-load to full load. V.R(%) = (No-load voltage-Full load voltage) ÷ Full load voltage=12-15÷15 = -20%.
88. Calculate the condition for maximum voltage regulation in the synchronous machine.
A. Φ=ϴs
B. Φ=2ϴs
C. Φ=4ϴs
D. Φ=8ϴs
Answer: A
Voltage regulation is maximum in the case of inductive load. The condition for maximum voltage regulation is when the power factor angle of the load becomes equal to the impedance angle.
V.R=Rp.ucos(Φ)+Xp.usin(Φ).
Differentiate V.R with respect to Φ and put it equal to zero.
We will get tan(Φ) = Xp.u÷Rp.u=tan(ϴs) then Φ=ϴs.
89. Zero voltage regulation can be only achieved in leading power factor load.
A. True
B. False
Answer: A
Zero voltage regulation only occurs during a leading power factor load. Condition for zero voltage regulation occurs when Φ+ϴs > 90o. For example – Capacitive load.
90. R.M.S value of the sinusoidal waveform v(t)=211sin(7.25t+7π÷78.3).
A. 149.19 V
B. 156.23 V
C. 116.57 V
D. 178.64 V
Answer: A
R.M.S value of the sinusoidal waveform is Vm÷2½ = 211÷2½ = 149.19 V and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vm.
