11. A 220 V, 1000 rpm, 60 A separately-excited dc motor has an armature resistance of .5 ω. It is fed from a single-phase full converter with an ac source voltage of 230 V, 50Hz. Assuming continuous conduction, the firing angle for rated motor torque at (-400) rpm is _________
A. 122.4°
B. 117.6°
C. 130.1°
D. 102.8°
Answer: D
During rated operating conditions of the motor
Eb = Vt-Ia×Ra = 220-60×.5=190 V.
As Eb=Kmwm so
Km=190×60÷(2×3.14×1000) = 1.8152 V-s/rad.
Back emf at (-400 rpm) is
Kmwm = 1.8152×(2×3.14×(-400))÷60 = -76 V. Now Vt = -76+60×.5 = -46 V.
The average voltage of a single-phase full converter is 2×Vm×cos(α)÷3.14.
The output of the converter is connected to the input terminal of the motor so α = cos-1(-46×3.14÷2×230×1.414) = 102.8o.
12. The unit of angular acceleration in rad/s2.
A. True
B. False
Answer: A
Angular acceleration is defined as a derivate of angular velocity with respect to time. It is generally written as α. The unit of angular velocity in rad/sec and time is second so the unit of angular acceleration in rad/s2.
13. Calculate the value of the angular acceleration of the motor using the given data: J= 50 kg-m2, load torque = 40 N-m, motor torque = 10 N-m.
A. -.7 rad/s2
B. -.6 rad/s2
C. -.3 rad/s2
D. -.4 rad/s2
Answer: B
Using the dynamic equation of motor J×(angular acceleration) = Motor torque – Load torque: 50×(angular acceleration) = 10-40 = -30, angular acceleration=-.6 rad/s2. The motor will decelerate and will fail to start.
14. The principle of step-up chopper can be employed for the ________
A. Motoring mode
B. Regenerative mode
C. Plugging
D. Reverse motoring mode
Answer: B
The step-down chopper is used in motoring mode but a step-up chopper can operate only in braking mode because the characteristics are in the second quadrant only.
15. A Buck-Boost converter is used to _________
A. Step down the voltage
B. Step up the voltage
C. Equalize the voltage
D. Step up and step down the voltage
Answer: D
The output voltage of the buck-boost converter is Vo = D×Vin ÷ (1-D. It can step up and step down the voltage depending upon the value of the duty cycle.
If the value of the duty cycle is less than .5 it will work as a buck converter and for a duty cycle greater than .5 it will work as a boost converter.
16. Which of the following converter circuit operations will be unstable for a large duty cycle ratio?
A. Buck converter
B. Boost converter
C. Buck-Boost converter
D. Boost converter and Buck-Boost converter
Answer: D
The output voltage of the buck converter and buck-boost converter are
Vo=Vin ÷ (1-D. and Vo = D×Vin ÷ (1-D) respectively.
When the value of the duty cycle tends to 1 output voltage tends to infinity.
17. Calculate the shaft power developed by a motor using the given data: Eb = 50V and I = 60 A. Assume frictional losses are 400 W and windage losses are 600 W.
A. 4000 W
B. 2000 W
C. 1000 W
D. 1500 W
Answer: B
Shaft power developed by the motor can be calculated using the formula P = Eb×I-(rotational losses) = 50×60 = 3000 – (600+400) = 2000 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.
18. Which one of the following devices has low power losses?
A. MOSFET
B. IGBT
C. SCR
D. BJT
Answer: C
SCR is a minority carrier device due to which it experiences conductivity modulation and its ON state resistance reduction due to which conduction losses are very low.
19. Servo motors are an example of which type of load?
A. Pulsating loads
B. Short time loads
C. Impact loads
D. Short time intermittent loads
Answer: B
Servo motors are motors with control feedback. The motor can be AC or DC. This is an example of short-time loads. They have a high torque to inertia ratio and high speed.
20. Load torque of the crane is independent of _________
A. Speed
B. Seebeck effect
C. Hall effect
D. Thomson effect
Answer: A
The Load torque of the crane is independent of speed. They are short time intermittent types of loads. They require constant power for a short period.
