1B. A 4 kVA, 200/400 V, 50 Hz, single-phase transformer gave the following test figures: No load (OC) (LV side) Short circuit (SC) (HV side) : 200 V, 0.7 A, 70 W : 15 V, 10 A, 80 W Sketch its equivalent circuit referred to the LV side. Also, find its efficiency and secondary terminal voltage on full load at unity power factor.

1B. A 4 kVA, 200/400 V, 50 Hz, single-phase transformer gave the following test figures: No load (OC) (LV side) Short circuit (SC) (HV side) : 200 V, 0.7 A, 70 W : 15 V, 10 A, 80 W Sketch its equivalent circuit referred to the LV side. Also, find its efficiency and secondary terminal voltage on full load at unity 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...

See similar textbooks

Similar questions

Can you solve D and E
A 1,000/200 V transformer takes 0.3 at p.f. of 0.2 on open circuit. Find the magnetising, and iron loss component of no- load primary current.
1. A 10 kVA, single-phase, 2400-V:120-V, 60 Hz transformer has an equivalent impedance referred to LV side of 0.01 + j0.09 Q. This transformer has a core loss of 100 W at rated voltage. When rated current is applied to the 2400-V winding, the exciting current in that winding is found to be 0.2 A. a. Find the full-load regulation of this transformer at a power factor of 0.8 lagging. b. Find the efficiency of this transformer at full load, 0.8 power factor lagging. c. If standard open-circuit and tests were performed on this transformer, what would be the measured values of these tests? 2. The following data were obtained on a 2.5 MVA, 50-Hz, 19.1-kV:3.81-kV single-phase transformer tested at 50 Hz: Voltage (V) Current (A) Power (kW) LV winding with HV terminals open-circuited: 3810 9.86 8.14 HV winding with LV terminals short-circuited: 920 141 10.3 a. Calculate the cantilever equivalent-circuit parameters as referred to the transformer high-voltage winding. b. Calculate the cantilever…
by High Voltage
6- A 600 kVA, 1-ph transformer has an efficiency of 92 % both at full- load and half-load at unity power factor. Determine its efficiency at 60 % of full load at 0.8 power factor lag.
Efficiency at full-load and half the full-load at 0.8 p.f. lagging current. Equivalent reactance referred to primary & referred to secondary..........ohm , ..........ohm Equivalent resistance referred to primary & referred to secondary..........ohm , ..........ohm Regulation of transformer at 0.5 p.f. lagging current The magnetizing current and reactance component of no-load are.........amper , .........ohm The working current and resistance component of no-load are.........amper , .........ohm *
1. A 10 kVA, single-phase, 2400-V:120-V, 60 Hz transformer has an equivalent impedance referred to LV side of 0.01 + j0.09 Q. This transformer has a core loss of 100 W at rated voltage. When rated current is applied to the 2400-V winding, the exciting current in that winding is found to be 0.2 A. a. Find the full-load regulation of this transformer at a power factor of 0.8 lagging. b. Find the efficiency of this transformer at full load, 0.8 power factor lagging. c. Ifstandard open-circuit and tests were performed on this transformer, what would be the measured values of these tests?
The maximum efficiency of a 200KVA, 6000/230volt, 50HZ, single phase transformer occurs at half load and is 95% at unity power factor. What is the full load efficiency at 0.7 power factor?
If a 50-turn transformer winding has a 120-V rms applied voltage, and if the peak coupling flux is 20 mWb, find the frequency of the applied voltage.
Topic- basic & const.
A 25 kVA, 2 kV(rms)/100 V(rms) single phase transformer is connected between a 120 V(rms) utility supply and a resistive load. The load current was measured to be 20 A(rms). i. Determine the active power of the load. ii. Determine the primary RMS current. iii. Determine the equivalent resistance of the transformer as seen by the utility supply.