1. In the rotor voltage injection method, when an external voltage source is in opposite phase with the main voltage then speed will ___________
a) Increase
b) Decrease
c) Remain unchanged
d) First increases then decrease
Explanation: In the rotor injection method, when an external voltage is in opposite phase with the main voltage net voltage decreases and the value of slip increases and the value of rotor speed decreases.
2. The rotor injection method is a part of the slip changing technique.
a) True
b) False
Explanation: Rotor injection method comes under slip changing technique. It uses an external voltage source to change the slip value. The load torque remains constant here.
3. The slip recovery scheme is a part of the synchronous speed changing technique.
a) True
b) False
Explanation: 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.
4. 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
Explanation: 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 of resistance.
5. 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
Explanation: 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.
6. 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
Explanation: At heavy loads, the current drawn is high due to which active power component increases. Increase in active power component increases the power factor of the machine.
7. At low values of slip, the electromagnetic torque is directly proportional to ___________
a) s
b) s2
c) s3
d) s4
Explanation: At low values of slip, the electromagnetic torque is directly proportional to slip value. Due to heavy loading slip value decreases which increases the ratio of R2÷s.
8. 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
Explanation: 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.
9. 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
Explanation: 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 case of the resistor so the angle between V & I is 90°. P=I2R=0×0×44=0 W.
10. The value of slip at which maximum torque occurs ________
a) R2÷X2
b) 4R2÷X2
c) 2R2÷X2
d) R2÷3X2
Explanation: 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.