Only expert should solve it 2:43 PM P & G.O+ " 40 221 28IMG-20240829-WA0029.jpg(388/2181)Questiona 5kva,200v/400v,50hz single phasetransformer gave the following test1.open circuit test Voc = 200v loc = 0.7A andPoc =60watt2.short circuit test V=22, current=16a,power=220w, calculate the equivalent ofthe transformer when refered to high voltagesideShare| |||EditоSet as||Deleten

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Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter30: Dc Motors

Section: Chapter Questions

Problem 6RQ: What is CEMF?

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A three phase full wave rectifier is shown below along with peak phase voltage Vm-169.7V. The load is purely resistive. The rectifier delivers Ipc = 100 A and the source frequency is 60 Hz. The DC output voltage is %3D VDc=280.7V and the output RMS voltage is equal to VRMS=280.93V. The efficiency, FF and RF are respectively equal to: Secondary D D D, R AD. Z D. D, Select one: O a. 99.83%, 100.08%, 4% O b. 99.83%, 55.08%, 4% O C. 99.83%, 100.08%, 16% O d. 87.83%, 100.08%, 4%
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Find VDc if VAN = VBN = VCN= 120 volts... Half-wave Three-phase Rectifier Conduction Waveform Periodic Time (T) A B 180 360 90° 450 540 time 270 VAN VeN VAN VEN Voc Vrc 30 120 150 270 390 510° time Output Voltage Waveform B
(a) Figure Q2 (a) shows the image from the digital osciloscope. It shows two sine waveforms, voltage (Ch1) and current (Ch2). The value of vertical (volt/div) for Ch1, (amp/div) for Ch2, and horizontal scales (time/div) are shown in the image. Analyze the figure and determine: iv) Peak to peak value and the frequency for each waveform (voltage, Ch1 and current, Ch2) v) Phase different between two waveforms. vi) If the vertical scale (volt/div) of Chl is change to 100 V. Determine the total of division of Chl that will be displayed by the oscilloscope. Ch1 Ch2 Ch1 50V Ch2 5A M5.00ms Figure Q2 (a)
A 60 Hz sinusoidal voltage with an RMS value of 120 V is applied to a single-phase half-wave SCR rectifier. The firing of SCR is delayed by 60 degrees. The DC output voltage of the rectifier would be: O A. 60 V O B. 90 O C. 81 V O D. 40 V O E. None of the other choices are correct
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A 60 Hz sinusoidal voltage is applied to an SCR full-wave rectifier. If the DC output voltage is 162 V and the firing angle of the SCR is 60 degrees, the peak voltage of the input sinusoidal would be: O A. 197 V O B. 228 V O C. Can't tell; need more information. O D. 340 V O E. None of the other choices are correct
Design of a DC-DC Converter. The Circuit Below shows a basic Buck-Boost Converter with a highly inductive Load. With a Deliberate mistake in the configuration of the Circuit Connection. The Value of L1 is 22mH, C1 is 10uF, R1 is 220 and L2 is 10mH. The circuit operates with a switching frequency of 20KHZ. The Duty cycle of the Clock signal V2 is given by the function below: ( 17 ) х0.2 Duty Cycle, k = 0. 55 + 100 Your Unique Duty Cycle = %3D Buck-Boost Converter Q1 IRG4BC10U D1 R1 222 1N4009 C1 LV1 12V L1 22mH 10µF L2 10mH a) Find the mistake in the configuration of the circuit and correct the connection error.
A single-phase full bridge uncontrolled rectifier is supplied by 220 V (rms) line voltage. The load is pure resistive. What is the output Ripple Factor ?
1) AC power in a load can be controlled by using a. Two SCR's in parallel opposition b. Two SCR's in series c. Three SCR's in series d. Four SCR's in series 2) The advantage of using free - wheeling diode in half controlled bridge converter is that a. There is always a path for the ac current independent of the ac line b. There is always a path for the dc current independent of the ac line c. There is always a path for the dc current dependent of the ac line d. There is always a path for the ac current dependent of the ac line 3) Silicon controlled rectifier can be turned on a. By applying a gate pulse and turned off only when current becomes zero b. And turned off by applying gate pulse c. By applying a gate pulse and turned off by removing the gate pulse d. By making current non zero and turned off by making current zero
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A single-phase full-wave rectifier is connected to an AC source with a peak voltage of 170 V and a frequency of 60 Hz. The load resistance is 502.

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