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- Appliances in home that use DC motor and briefly explain why it works as reverse DC generatorDifferentiate between the AC and DC magnetic fields applied to an induction motor stator.Maz a) A sinewave is having a frequency of 50 Hz and instantaneous value of 250 Volts at 1.3 milli seconds. Find D The maximum value of sinusoidal voltage waveform and i) The value of voltage at 2.1 milli seconds. b) Determine the instantaneous equation for the voltage waveform shown in figure, if Vp=100 Volts, Time period is 10 milli seconds and 0 =40 degrees. on Vp Upload your Answers steps in the "Final Answer Submission" Link provided in the Moodle. a) (i) Maximnum value = a) (ii) Value of voltage at 2.1 ms = b) Angular frequency =
- The back emf of a dc motorThe DC shunt generator with the terminal voltage 300 V, armature drop 8 V and the each brush drop 2 V, then the value of generated emf is1. The operation of Commutator in DC Motor is * O a) to convert dc to ac O b) to convert ac to dc O c) to convert fixed dc to variable dc O reverse the current direction in the rotating windings each half turn
- Draw occ at 800 rpm At 1000 rpm And 600 rpm and air gap Field Armature emf V Armature emf V Armature emf V current(A) At 800rpm At 1000rpm At 600rpm 8 10 6 0.3 28 35 21 0.5 46 57.5 34.5 1.0 94.5 118.125 70.875 1.5 143 178.75 107.25 2.0 175 218.75 131.25 2.5 195.5 244.375 146.625 3.0 211 263.75 158.25 3.5 224 263.75 168 4 234 280 175.875 4.5 243 293.125 182.625 5 252 314.375 188.625 6 266 331.875 199.125 7 277 347.5 207.375 8 286 368.125 214.5 9 295 368.125 220.875 10 302 226.5 226.5 The drawing must be handwritten without using a computer scale X-axis 1cm = 1 y-axis 1cm =20At no load, a 250V shunt motor runs at 1000 rpm and draws 8A. Armature and field resistances are 0.20 ohm and 250 ohms, respectively. If the motor draws a current of 50A, its speed will be8- The frequency of induction motor is 50 Hz and the slip is 2.5% then frequency of rotor current is a. 1.25 HZ b. 1.5 HZ c. 1.2 HZ d. 1 Hz
- A field excitation of 20 A in a certain alternator results in an armature current of 400 A in short circuit and a terminal voltage of 2000 V on open circuit. The magnitude of the internal voltage drop within the machine at a load current of 200 A isA generator is feeding the load. A second load will then be connected in parallel with the first load. The generator has an idle frequency of 61Hz and an Sp slope of 1MW/Hz. Load2 draws 800kW active power at 0.707 reverse power factor, while load1 draws 1000kW active power at 0.8 reverse power factor. What will be the operating frequency of the system after load 2 is connected? *Please explain the solution step by stepThe frequency of rotor current in an induction motor is slip times the frequency of stator current. slip times the frequency of supply, One by slip times the frequency of stator current. One by slip times the frequency of supply
