number 1 only please thank you The power output is(75 x103) = (0.79)(75 x 10) W326And so(0.79 )(75 x 103Nmax == 97.53%(0.79)(75 x 10')+750+750138Assessment Task 61. A quantity is expressed in per unit if it is divided by a chosen base quantityhaving the same physical dimension). Suppose that for a 10-kVA,2400-V/240-V transformer we chooseP_(base )=10kW v_[1 base )=2400v v_(2 base )=240VThis transformer has the following test dataopen-circuit test (on low-voltage side):short-circuit test (on high-voltage side):Convert all test data into per-unit values and find the series equivalent240 V, 0.8 A, 80 W80 V, 5.1 A, 220 wresistance in per unit.2. The results of open-circuit and short-circuit tests on a 25-kVA, 440-V/220-V,60-Hz transformer are as follows:Open-circuit test. Primary open-circuited, with instrumentation on the low-voltage side. Input voltage, 220 V; input current, 9.6 A; input power, 710 w.Short-circuit test. Secondary short-circuited, with instrumentation on the high-voltage side. Input voltage, 42 V; input current, 57 A; input power, 1030 W.Obtain the parameters of the exact equivalent circuit (Figure. 2-4), referred to thehigh voltage side. Assume that R, = a'R:3. A 110-kVA, 2200-V/ 110-V, 60-Hz transformer has the following circuit constants:R1= 0.22 0, R; = 0.5 mo, X, = 2.00, X2 = 5 m0, Rc = 5.5 ko, and Xm = 1.1 kn. Duringand X1= a'X2 voltage sideone day (24 hours) the transformer has the following load cycle: 4 h on no-load;8h on % full-load at 0.8 power factor; 8 h on 14 full-load at unity power factor; and; and 4 h on full-load at unity power factor. Assuming a constant core loss of1.346 kW. Find the all-day efficiency of the transformer4. The ohmic values of the circuit parameters of a transformer, having a turns ratio of5, are R: = 0.5 0; R2 = 0.021 0; X: = 3.2 0; X2 = 0.12 Q; Rc = 350 n, referred tothe primary; and Xm= 98 0, referred to the primary. Draw the approximateapproximate equivalent circuits of the transformer, referred to (a) the primary and(b) the secondary.. Show the numerical values of the circuit parameters. Thecircuits are respectively shown in Figure. 2-12(a) and Figure. 2-12(b). The calculations areas follows: (a) R', X', Rc', X'm (b) R", X", R"c. X"m (c)102s amAR A(a)(b)Figure. 6.12 Circuits

Given, Base power= 10 KW Base Voltage (HV)=2400 V Base voltage (LV)=240 V Base current (LV)=Base…

1. A quantity is expressed in per unit if it is divided by a chosen base quantity having the same physical dimension). Suppose that for a 10-kVA, 2400-V/240-V transformer we choose P_(base )=10kW V_(1 base )=2400V V_(2 base )=240V This transformer has the following test data open-circuit test (on low-voltage side): short-circuit test (on high-voltage side): Convert all test data into per-unit values and find the series equivalent resistance in per unit. 240 V, 0.8 A, 80 W 80 V, 5.1 A, 220 W

1. A quantity is expressed in per unit if it is divided by a chosen base quantity having the same physical dimension). Suppose that for a 10-kVA, 2400-V/240-V transformer we choose P_(base )=10kW V_(1 base )=2400V V_(2 base )=240V This transformer has the following test data open-circuit test (on low-voltage side): short-circuit test (on high-voltage side): Convert all test data into per-unit values and find the series equivalent resistance in per unit. 240 V, 0.8 A, 80 W 80 V, 5.1 A, 220 W

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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