Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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- 1. A motor load consists of a resistance of 6 Ohms in series with an inductance of 12 mH. Assume 120Vac, 60 Hz supply. a. What is the complex impedance of the load? b. What is the ac current through this load? c. What is theta, the angle between voltage and current through this load? d. What is the power factor? e. What capacitance should be added in parallel with this load to correct the power factor to 1? f. What is the current from the supply when the power factor is corrected?arrow_forwardQuestion 4 For the circuit shown, what is the complex power, S, consumed by the load impedance? (Assume the voltage source is in RMS.) a 15 N V = 150 20° V 150 mH W = 300 rad/s O 1.58/18.4° kVA O 1.35/25.5° kVA O 530.3/45.0°VA O 474.3/71.6°VA Question 5 124arrow_forwardCalculate the apparent power, real power, andreactive power for the circuit shown in Figure.Draw the power triangle. Assume a.) f = 50 Hz. b.) f = 0 Hz (DC)arrow_forward
- For the circuit shown, what is the average power, Py, consumed by the 1+j5 2 load impedance? (Assume the current source is in RMS.) 2Ω 10 Is = 3 20° A -j10 j5 0 19.04 W O 8.70 W 1.80 W O 0.84 W wwarrow_forwardFor the following voltage and current phasors, calculate the complex power, apparent power, real power, and reactive power. Determine whether the power factor is leading or lagging. V = 300/30° V rms, I = 0.5/60° A rms The value of apparent power is The value of real power is The value of reactive power is The power factor is (Click to select) ✓ W. VA. VAR.arrow_forwardTwo loads connected in parallel take a total of 2.4 kW with a lag fp of 0.8, from a line at 120 V rms and 60 Hz. of the loads absorbs 1.5kW with lagged fp of 0.707. Deter- mine: a) the fp of the second charge, b) the element in parallel required to correct the fp of the two charges and convert it to behind 0.9.arrow_forward
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