What is a disadvantage of starting with autotransformer coils rather than with resistors?
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What is a disadvantage of starting with autotransformer coils rather than with resistors?
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- A Flyback converter (for questions 1-3) shown below has the following parameters. Vin-100V, Vo-20v, Po-40W, D-0.25, Fsw-250kHz 1. What is the turns ratio N1:N2 between the primary side winding and the secondary side winding? A. NI: N2 - 5: 3 B. NI: N2 - 3:5 C. NI: N2- 5:2 D. None of the above9-In a the shunt field winding is connected in parallel with the armature winding only. Ish Series . Field EJR Shunt Rsh Rse Field 10-The consist of and 11- E = (V – IaRa ) then N = LOADL-1). The 3-phase wound-rotor induction motor has .... . a) high starting torque b) low starting current c) both "a" and "b" d) neither "a" nor "b" L-2). The 3-phase wound-rotor induction.........a) motor has a wire wound rotor b) motor's rotor can be connected to an external resistive load by way of slip rings and brushes c) motor requires more maintenance than the 3-phase squirrel-cage induction motor d) all of the above L-3). The 3-phase wound-rotor induction motor's rotor circuit's resistance should be adjusted so it is__________ during starting. a) low b) high c) either "a" or "b" d) neither "a" nor "b" L-4). Will a 3-phase wound-rotor induction motor start with no external circuit connected to its rotor leads? a) yes b) no
- Two identical 3-phase, 500-V induction machines are mechanically coupled for a back-to-back test. One machine is motoring from a 500-V, 50-Hz supply and the other is generating on to a 3-phase, 500-V synchronous machine for which the fre- quency can be controlled. It is required to run the motoring induction mạchine at 40 hp. Neglecting the slight change in the generating machine parameters due to oper- ating at a reduced frequency, estimate the rating of the synchronous machine for this duty and the frequency at which it must be run. Line input readings for each machine No load 500 V 8.5 A 1-6 kW Locked rotor 100 V 30 A 1-6 kW Friction and windage loss per machine 0-8 kW. R = R2'A Flyback converter (for questions 1-3) shown below has the following parameters. Vin=100V, Vo=20v, Po=40W, D=0.25, Fsw=250kHz iin i out in 1. What is the turns ratio N1:N2 between the primary side winding and the secondary side winding? A. Ν1: N25:3 B. N1: N2 = 3:5 C. N1: N2= 5:2 D. None of the above %3D 2. What is the maximum voltage on the power switch? A. 120V В. 133V С. 100V D. None of the above 3. What is the maximum voltage on the diode? Α. 40V В. 133V C. 80V D. None of the above 4. The magnetizing inductor at the primary side winding is 100uH, what is the peak current in the power switch? Α. 1.8Α В. 1.6A С. 2.1A D. None of the aboveQ5. (A) Select the correct answer for the alternatives given 1-In transformer, the load current at maximum efficiency is equal to rated load current multiplied by a) Peu Pa Pru Pi 2- When a series de motor is operated without a load, which of the following conditions occur? b) The voltage requirement increases d) The armature speeds out of control a) The armature draws excessive current c) The armature will not turn P₁ d) Peu 3- A 220V shunt motor develops a torque of216 N.m at an armature current of 40 A. The torque developed when the armature current is 20 A b)108 N.m c) 864 N.m a) 432 N.m d) 54 N.m 4-A single phase transformer has maximum efficiency at 75% of full load. Its copper loss at full load should be a) equal to core loss, b) less than core loss. c) more than core loss, d) equal to 75% of core loss 5- The efficiency of a transformer is 98% at full-load with 0.8 lagging power factor, what will be the efficiency when supplies full-load with 0.8 leading power factor? c) more…
- a. Explain the terms, coil span, coil pitch, short pitching and chording of coils.b. Equalizer rings are needed in lap wound dc machines not in wave wound explain the reason andthe use of equalizer rings with the help of an example.c. What are dummy coils?d. What is the purpose of using short-pitched coils in ac windings?A Flyback converter (for questions 1-3) shown below has the following parameters. Vin=100V, Vo=20V, Po=40W, D=0.25, Fsw=250kHz V in 1. What is the turns ratio N1:N2 between the primary side winding and the secondary side winding? A. N1: N2 = 5: 3 B. N1: N2 = 3:5 C. N1: N2= 5:2 D. None of the above 2. What is the maximum voltage on the power switch? A. 120V В. 133V C. 100V D. None of the above 3. What is the maximum voltage on the diode? А. 40V В. 133V С. 80V D. None of the above 4. The magnetizing inductor at the primary side winding is 100uH, what is the peak current in the power switch? A. 1.8A В. 1.6A С. 2.1А D. None of the aboveWhat is a single - phase induction motor? How does a single- phase induction motor work? How do you test a single phase motor? Which capacitor is used in single phase motor? What are the advantages of single phase induction motor? What are the applications of single phase induction motor? How does an induction motor start? What are the differences between single and three phase I.M ? What are the types of induction motor? What is the difference between induction motor and normal motor? Why three-phase induction motor self start while single phase motor does not self start?
- 14.13 For controlling the speed of de motor of 150 hp rating, the following types of converters are normally used (a) single-phase full converters (b) single-phase dual converters (c) three-phase full converters (d) three-phase dual convertersA 54-slots, 216 segments has four poles lap armature windings that is equalized 100 per cent at the commutator. a) How many equalizer connections are there? b) To what two segments is the first equalizer be connected?A freewheeling DC motor is driven with 1000 rpm and a nominal load torque of 80Nm. The field circuit resistance is 200 and the armature circuit resistance is 0.2 O. The field coils are fed with a zero degree transmission angle through a single-phase fully controlled converter connected to a single-phase 240V source. The armature circuit is fed from a single phase 400V source through another fully controlled converter. Magnetic saturation has been neglected, the motor constant is given as 0.8 V - s / A - rad. Assuming that there is no fluctuation in armature and field current, find the control angle of the converter feeding the armature at nominal load. If you want to leave the question blank, click again on the option you marked.