5. A 3-phase, 50 Hz, star - connected, 2000 - KVA, 2300-V alternator gives a short circuit current of 600 A for a certain field excitation. With the same excitation, the OC voltage was 900 V. The resistance between a pair of terminals was 0.12 . Find the full load regulation at (a) UPF (b) 0.8 pf lagging and (c) 0.8 pf leading [7.3% ; 23.8 %; -13.2%]

5. A 3-phase, 50 Hz, star - connected, 2000 - KVA, 2300-V alternator gives a short circuit current of 600 A for a certain field excitation. With the same excitation, the OC voltage was 900 V. The resistance between a pair of terminals was 0.12 . Find the full load regulation at (a) UPF (b) 0.8 pf lagging and (c) 0.8 pf leading [7.3% ; 23.8 %; -13.2%]

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

6) Consider the circuit in Figure #2 that was set as an example at the beginning. Use different values of 61, 63, R₁, and R3 that you will choose arbitrarily-I am expecting that each one of you will have your own set of values (no plagiarism and find the three electric currents (values in A and their directions). b R 2 Figure#2 X y с R RA d u g f
Plz correct solution.....
Needs Complete typed solution with 100 % accuracy.
A PMMC instrument gives 19 mA at full scale reading when a potential difference across its terminals is 41mV. Show that how it can be used (a) as an ammeter for the current measurement in the range of 0-2 A (b) as a 0 - 630V range voltmeter for the voltage measurement. (c) find the multiplying factor of shunt and (d) voltage amplification Rsh shunt resistance R1 series resistance multiplying factor for current voltage amplification
A battery with E = 3.00V and no internal resistance supplies current to the circuit shown in the following figure: (Check image da_2 for the figure) When the double position switch S is open, as shown, the current in the battery is 1.01 mA. When the switch is closed in position A, the battery current is 1.28 mA. When the switch is closed in position B, The current in the battery is 2.04 mA. Determine the resistors: A) R1= B) R2= C)R3= present results in kOhms
The circuit on the right has several resistors and ca- pacitors, as well as an ideal ammeter. The battery is assumed to have negligible internal resistance. Right after the switch is closed: a) What current I₁ does the ammeter read? b) What power P₁ is being dissipated by the 50.0 2 resistor? What charge Q₁ is stored on the 10.0 µF ca- pacitor? After a long time has passed: c) What current I2 will the ammeter read? d) What power P₂ will the 50.0 2 resistor dissipate? What charge Q2 will be stored on the 10.0 μF capacitor? 100.0 V 25.0 Ω |75.0 Ω 15.0 με www 20.0 μF 50.0 ΩΣ www 15.00 25.0 Ω 10.0μF $25.0 2
b) The unknown resistance is determined by Wheatstone bridge and is given by Rx = R. where R; = 90 = 0.5% 2 R; = 900 = 0.8% 2 R; = 825 = 0.6% 2 Determine the magnitude of the resistance and the limiting error in % and ohms for the unknown resistance R.
provide implicit general solution, show all steps please
In a two-terminal network, the open-circuit voltage measured at the given terminals by an electronic voltmeter is 100 V. A short-circuit current measured at the same terminals by an ammeter of negligible resistance is 5A. If a load resistor of 8092 is connected at the same terminals, then the current in the load resistance will be
How can I simplifie this circuit ?
TOPIC: Combination Circuit
Handwrite and step by step solution