This set of Tough Electric Drives Questions and Answers focuses on “Dynamics – Load Torques that Vary with Angle of Displacement of the Shaft”.
1. Duty cycle (D) is _______
b) Ton÷(Ton+ Toff)
c) Ton÷2×(Ton+ Toff)
Explanation: Duty cycle (D) is defined as the ratio of time for which system is active to the total time period. It is also known as the power cycle. It has no unit.
2. A 220 V, 1000 rpm, 60 A separately-excited dc motor has an armature resistance of .5 ω. It is fed from 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 _________
Explanation: 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. Average voltage of 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.
3. The unit of angular acceleration is rad/s2.
Explanation: 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 is rad/sec and of time is second so the unit of angular acceleration is rad/s2.
4. 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
Explanation: 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.
5. The principle of step-up chopper can be employed for the ________
a) Motoring mode
b) Regenerative mode
d) Reverse motoring mode
Explanation: The step-down chopper is used in motoring mode but a step-up chopper can operate only braking mode because the characteristics are in the second quadrant only.
6. 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
Explanation: 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 duty cycle greater than .5 it will work as a boost converter.
7. 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
Explanation: 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.
8. 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
Explanation: 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.
9. Which one of the following devices have low power losses?
Explanation: SCR is a minority carrier device due to which it experiences conductivity modulation and it’s ON state resistance reduction due to which conduction losses are very low.
Sanfoundry Global Education & Learning Series – Electric Drives.
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