>- The open circuit and short circuit tests on a 10 KVA, 125/250 V, 50 Hz single phase transformer gave the following results: O.C. Test: 125 V, 0.6 A, 50 W (on L.V. side) S.C. Test: 20 V, 40 A, 177.78 W (on H.V. side) Calculate: (i) Magnetizing current (I). *** (ii) The voltage regulation on full load, 0.9 leading p.f.

>- The open circuit and short circuit tests on a 10 KVA, 125/250 V, 50 Hz single phase transformer gave the following results: O.C. Test: 125 V, 0.6 A, 50 W (on L.V. side) S.C. Test: 20 V, 40 A, 177.78 W (on H.V. side) Calculate: (i) Magnetizing current (I). *** (ii) The voltage regulation on full load, 0.9 leading p.f.

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter28: Three-phase Transformers

Section: Chapter Questions

Problem 4RQ: An open-delta three-phase transformer system has one transformer center-tapped to provide a neutral...

See similar textbooks

Similar questions

An open-delta three-phase transformer system has one transformer center-tapped to provide a neutral for single-phase voltages. If the voltage from line to center tap is 277 V, what is the high-leg voltage?
Rework Example 3.12 for a+10 tap, providing a 10 increase for the high-voltage winding.
A 10-kVA, 200-V/400-V, 50-Hz, single-phase transformer gave the following test results : OC test (HV winding open) : 200 V, 1.3 A, 120 W. SC test (LV winding shorted) : 22 V, 30 A, 200 W. Calculate (a) the magnetizing current, and (b) the equivalent resistance and leakage reactance as referred to the low voltage side.
A single phase transformer gave the following results: Open Circuit Test: 250V, 1A, 80W Short Circuit Test: 20V, 12A, 100W Calculate the parameters the we need for both tests
19.A 25kVA, 2400/240, 60Hz single phase transformer gave the following short circuit test result Vsc = 100V, Isc = 10.41A, Psc = 116. Find the equivalent resistance in ohms referred to primary. Assume Ip rated is equal to Isc. round off answer to 4 decimal places if necessary.
Solve the following problems with complete solutions and units usingtwo(2) decimal digits.3. A 100-kVA, 6600/330-V., 50 Hz, 1-phase transformer took 10 A. and436 W. at 100 V. in a short-circuit test, referring to the high voltageside. Calculate the voltage to be applied to the high voltage side onfull-load at power factor of 0.8 lagging when the secondary terminalvoltage is 330 V.
A 75KVA, 230/115V, 60 Hz transformer was tested with the following result: Short Circuit: 9.5V, 326A, 1200W Open Circuit: 115V, 16.3A, 750W Determine the (a) equivalent impedance, reactance and resistance in per unit and (b) voltage regulation at rated load, 0.8 pf lagging.
The efficiency of a 20 kVA 15.4 / 0.4 kV three-phase power transformer is 79.1% for a load with unit power factor at rated power. Transformer Relative short circuit voltage% Uk = 30.0 and relative idle current% I0 = 30. Since the transformer is connected star-delta; Note: In 3-phase electrical machines, the current-voltage information given on the label is always line values. Phase value When given, it must be specified separately. Therefore, all current-voltage information given in the question are line values. 1-Draw the primary reduced equivalent circuit. 2-What is the power value measured in the short circuit test of this transformer? -The value of iron losses of this transformer what?
No-load operation and short circuit tests were performed on three-phase, 50 Hz 480/220 delta / Y connected transformer. In these testsThe line voltage, line current and three-phase total power have been measured from the high voltage windings and the test results are below.pictures. 100 A DC current when 1.08 V DC voltage from non-transferable software is applied to the low voltage windingshas attracted.Find the single phase equivalent circuit parameters of this transformer.idle test:Voltage: 480V Power:114w current: 0.71Ashort circiut test:V: 10v P: 78w c: 18.3a
A single-phase 1,000 VA, 230/115 V transformer has been tested to determine its equivalent circuit. The results of the tests are shown in Table. 1. All the data given in the Table. 1 were taken from the primary (high voltage) side of the transformer. a) Find the equivalent circuit of this transformer referred to the low-voltage side of the transformer b) Neglecting magnetizing impedance, find the transformer’s voltage regulation at rated conditions andi) 0.8 power factor laggingii) 1.0 power factoriii) 0.8 power factor leading c) Including magnetizing impedance, determine the transformer’s efficiency at rated conditionsand 0.8 power factor lagging. Table 1. Open-circuit test Short-circuit test VOC = 230 V IOC = 0:45 A POC = 30 W VSC = 19:1 VISC = 8:7 APSC = 42:3 W
Three single-phase, 50 kVA, 2300/230 V, 50 Hz transformers are connected to form a 3-phase, 2300/230V transformer bank. The equivalent impedance of each transformer referred to low voltage is 0.03 + j0.02 2. The 3-phase transformer supplies a 3-phase load, 120 kVA, 230 V, 0.9 p.f (lag). Determine (a) the transformer winding currents. (b) the voltage regulation. (comment on the obtained results)
The efficiency of a 20 kVA 15.4 / 0.4 kV three-phase power transformer is 74,1% for a load with unity power factor at rated power. Relative short circuit voltage% Uk = 30.0 and relative no-load current% I0 = 30. Since the transformer (Star-delta) is connected; Note: In 3-phase electrical machines, the current-voltage information given on the label is always line values. Phase valueWhen given, it must be specified separately. Therefore, all current-voltage information given in the question are line values. 1-) Draw the primary reduced equivalent circuit. 2-) What is the power value measured in the short circuit test of this transformer? 3-) What is the iron loss value of this transformer ?