The three loads in the circuit seen in (Figure 1) are S₁ =4+1 kVA, S2 5+ 12 kVA, and S₁ 10+50 kVA. Suppose that v₁ =v2 = 145 V. Calculate the complex power associated with Vg1. Use positive value for absorbed power and negative value for delivered power. Express your answer in kilovolt-amperes to three significant figures. Enter your answer in rectangular form. View Available Hint(s) Hint 1. How to approach the problem The complex power associated with a source can be found from the voltage of the source and the current of the source. In order to find consider the currents into which it splits and determine them from the given power values. After finding the currents, you may calculate the source by applying KVL Figure 1 of 1 Sgl 19 ΑΣΦ Η VOC 0.05 w Submit Previous Answers Request Answer 0.15 w S₁₁V(rms) S₁ X Incorrect; Try Again SV(rms) 0.05 0 w Part B ? kVA

The three loads in the circuit seen in (Figure 1) are S₁ =4+1 kVA, S2 5+ 12 kVA, and S₁ 10+50 kVA. Suppose that v₁ =v2 = 145 V. Calculate the complex power associated with Vg1. Use positive value for absorbed power and negative value for delivered power. Express your answer in kilovolt-amperes to three significant figures. Enter your answer in rectangular form. View Available Hint(s) Hint 1. How to approach the problem The complex power associated with a source can be found from the voltage of the source and the current of the source. In order to find consider the currents into which it splits and determine them from the given power values. After finding the currents, you may calculate the source by applying KVL Figure 1 of 1 Sgl 19 ΑΣΦ Η VOC 0.05 w Submit Previous Answers Request Answer 0.15 w S₁₁V(rms) S₁ X Incorrect; Try Again SV(rms) 0.05 0 w Part B ? kVA

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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Question oneA balanced star connected load comprises of 20 ohm resistance and 0.07 Henry inductance in each branch. If the load supplied with 415V at 50Hz, calculatea) the line currentb) the power factorc) the total power supplied to the load PLEASE FOLLOW THIS APPROACH TO SOLVE THE QUESTION. THANK YOUStep OneDeduce the inductive reactance.STEP TwoWrite down a rectangular expression for the impedance per branch.STEP ThreeConvert the rectangular expression into polar expression.STEP FourBecause impedance will be involved in the analysis deduce the phase voltage.STEP FiveSolve for the phase current and hence the line current.STEPVSixUse the phase angle of the current to deduce the power factor. STEP SevenFind the apparent power.
Suppose there is no power lost in the wires used to distribute power in a typical home. The following voltages and currents are calculated for thecomponents:va=120 V ia=−10 Avb=120 V ib=10 Avf=−120 V if=3 Avg=120 Vvh=−240 V ih=−7 AIf the power in this modified model balances, what is the value of thecurrent in component g?
In the circuit given in Figure 2 vs(t)=√2.50 cos(4t+60°) a) Calculate the equivalent load impedance. Indicate whether the load is inductive or capacitive. B) Calculate the load current load. Write the expression for the time dependent i(t). C) Calculate the load voltage. Write the expression for time dependent v(t) d) Calculate the active, reactive and apparent power of the load. Determine the power factor Y=3
Given: Gen 1 1500 KVA, 3Phase , 2.3 KV, 4 Poles No load speed= 1830 rpm Speed @ 1500 Kw = 1788 rpm Gen 2 5000KVA, 3 phase, 13.2 KV ,2 Poles No load Speed = 3684 rpm Speed @ 5000 Kw = 3552 rpm If the Total Load connected is 6000 Kw, A. The Power supplied by Generator 1 is... 1590.068 Kw 1690.068 Kw 1631.068 Kw 1531.068 Kw B. The power supplied by Generator 2 is.... 5390.982 Kw 4390.982 Kw 5368.982 Kw 4368.982 Kw C. The common Frequency of Generator 1 and generator 2 is .... 59.477 Hz 60.00 Hz 58.477 Hz 58.70 Hz D.The Speed of generator 1 is ...... 1720.35 RPM 1784.35 RPM 1700.35 RPM 1710.35 RPM E. The Speed of generator 2 is.. 3510.68 RPM 3520.68 RPM 3568.68 RPM 3500.68 RPM Please all subparts in short...
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Assume that vg = 120 V rms. Finda. The secondary current.b. The installation efficiency Pload/Psource.c. The value of the load resistance which can absorbthe maximum power from the given source.
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A three-phase 240-volt table saw requires 36 kW. Calculate the load current for the table saw. The current and the voltage are out of phase by an angle of 32 degrees. Hint 1: 30kW here refers to working power, not apparent power. Hint 2: The equation for working three-phase power is found in Eq. 12.21. Hint 3: Refer to Figure 12.12 about calculating theta
power system has a base power rating of 154 kVA and a base impedance of 24ohm. The system is experiencing a bit of a power surge and is current running at I = 0.85pu V = 0.7pu Z = 0.72 pu, and S= 1.2 pu. What are the actual values of the current, voltage, impedance, and power of the system?
Given: Gen 1 1500 KVA, 3Phase , 2.3 KV, 4 Poles No load speed= 1830 rpm Speed @ 1500 Kw = 1788 rpm Gen 2 5000KVA, 3 phase, 13.2 KV ,2 Poles No load Speed = 3684 rpm Speed @ 5000 Kw = 3532 rpm If the Total Load connected is 6000 Kw, The Speed of generator 2 is.. A. 3510.68 RPM B. 3520.68 RPM C. 3568.68 RPM D. 3500.68 RPM
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It is required to know 1. The Real Power of the Generator: 2. The Reactive Power of the Generator 3. The Power Factor of the Generator 4. The Reactive power necessary to improve the power factor in the point "x" at 0.95. LINE 1 P 1 KW consuming Q=1 KVAR consuming LOAD 1 P5 LINE 2 LOAD 2 KN consuming Q =9.45 KVA Rewatering Z=0.2236 angle 63.43 Q P triphasic 9 KW consuming Q3o LOAD 3 12 KVAR consuming Three-phase P = 30 KW consuming pf = 0.8 VLL 208 angle 30°v