and a full-load copper loss of 460 W. (a) The above transformer is to be connected as an auto transformer to supply a load at 150 V from 250 V source. Please draw the auto transformer connection diagram. And what kVA load can be supplied without exceeding the current rating of the windings? (For this part assume an ideal transformer.) (b) Find the efficiency with the kVA loading of part (a) and 0.85 power factor.

and a full-load copper loss of 460 W. (a) The above transformer is to be connected as an auto transformer to supply a load at 150 V from 250 V source. Please draw the auto transformer connection diagram. And what kVA load can be supplied without exceeding the current rating of the windings? (For this part assume an ideal transformer.) (b) Find the efficiency with the kVA loading of part (a) and 0.85 power factor.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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It belongs to the no-load (no-load) operation of the 36/240 V phase transformer and a phase measured from the short-circuit test. values are given in the table below. High voltage by applying rated voltage to the low voltage winding (The winding ends were left em pty and the unloaded test results were obtained. Rated currents to high voltage gates The decreasing energies were obtained in a short time from a short circuit by applying tests In direct current test The transformer with 1 V direct voltage applied to the low voltage winding draws 500 A direct current. All equivalent Calculate the circuit elements moved to the low voltage side. Boşta çalışma testi Güç Gerilim Akım 36 V 68 W 2,8 A Kısa devre testi Gerilim Güç Akım 15 V 325 W 42,2 A |
TRUE or FALSE: If a transformer is loaded from full-load value to half-load value, then the amount of core loss is also halved.
4. 100 kVA, 6800-240 V, 50 Hz single phase step down transformer has the following parameters: RLs=0.0082 RHs=2.62 Rfe-HS=50002 XLs=0.015N XHs=5.2N Xm.Hs=88002 Transformer is delivering 80% of rated load at rated voltage and 0.95 lagging power factor. Determine the following: a) The equivalent impedance of the transformer referred to primary. b) Input impedance of the combined transformer and load. c) Actual input voltage at HV side. d) Input impedance if load is disconnected. e) Exciting current at no load.
Find the elements of the equivalent circuit attributed to the secondary of a single-phase transformer. The no-load voltages in the primary and secondary circuit are 200 volts and 400 volts, respectively, with a frequency of 50 Hz. The primary no-load current at the voltage of 200 volts is 0.7 amperes and the loss of the core is equal to 70 watts. The resistance and reactance of the primary are 3.2 ohms and 1.25 ohms, respectively. The resistance and reactance of the secondary are 0.6 ohm and 0.07 ohm, respectively. Draw the equivalent circle, showing the values of its elements.
What will happen if proper polarity is taken for granted in paralleling transformer? If voltages are different? Which is much more dangerous?
30-kVA, 2200/220-V, 50-Hz distribution transformer is tested for efficiencyand regulation as follows :O.C. test( Low voltage side ) : 220 V 4.2 A, 148 W S.C. test ( Low voltage side ) : 86 V 10.5 A, 360 W 1-The working current and resistance component of no-load are.........amper , .........ohm * 2-Efficiency at full-load and half the full-load at 0.8 p.f. lagging current. * 3-Equivalent resistance referred to primary & referred to secondary..........ohm , ..........ohm * 4-Regulation of transformer at 0.5 p.f. lagging current * 5-Equivalent reactance referred to primary & referred to secondary..........ohm , ..........ohm * 6-The magnetizing current and reactance component of no-load are.........amper , .........ohm *
A 36/240 V phase transformer, belonging to a phase measured from the no-load (idle) run and short-circuit test. values are given in the table below. High voltage by applying rated voltage to the low voltage winding (The winding ends were left unloaded and the results of the no-load test results were obtained. By applying nominal current to the high voltage windings, the low voltage windings were short-circuited, and the results of the short circuit test were obtained. In direct current test 1 V direct voltage is applied to the low voltage winding, and the transformer draws 500 A direct current. All equivalent Calculate the circuit elements as moved to the low voltage side. Idle run test: Voltage : 36V Power : 68W Current : 2,8A Short circuit test: Voltage : 15V Power : 325W Current : 42,2A
4. Why the efficiency of the transformer is high at half load compared to full load?
High voltage engineering