This is a part of a review I'm studying, NOT a graded assignment, please do not reject. Thank you!I only want the solution for the 4th question that the black arrow is pointing to!The answers for the first and third question are given below; I dont think the solution to the second question is necesarry to solve the 4th:S1 = 1510.6 at 3.69 degrees MVAS2 = 1510.6 at -26.3 degrees MVAS1 p.u. = 1.51 at 3.69 degrees p.u.S2 p.u. = 1.51 at -26.3 degrees p.u. Circuit analysis using phasor/impedance, complex power calculation: A three-phase RLC circuitis shown below. The reactance of the capacitor is 40 Ohm. The reactance of the inductor is 80 Ohm, and theresistor is 8 Ohm. The sending end voltage is v₁ (t) = √2³45 cos(wt + 7) kV, where w = 27 · 60 rad/s. Thereceiving end voltage is v2(t) = √√2³45 cos(wt) kV. The power base is 1000 MVA and the voltage base (L-L) is345 kV.Please compute total transferred complex power S₁ and S₂ in the physical unit.Please draw a per-phase phasor/impedance diagram using per unit values.Please compute the per unit complex power S₁ and S₂.Find out the real power loss by the resistor, reactive power consumption by the inductor and the reactivepower generated by the series capacitor in per unit.S₁→>>>>RLсqoldinguFigureVr

For a three-phase RLC circuit, the following information is given:base parameters: Sb=100 MVA,…

Answered: Circuit analysis using…

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

Related questions

Question

This is a part of a review I'm studying, NOT a graded assignment, please do not reject. Thank you!

I only want the solution for the 4th question that the black arrow is pointing to!

The answers for the first and third question are given below; I dont think the solution to the second question is necesarry to solve the 4th:

S1 = 1510.6 at 3.69 degrees MVA

S2 = 1510.6 at -26.3 degrees MVA

S1 p.u. = 1.51 at 3.69 degrees p.u.

S2 p.u. = 1.51 at -26.3 degrees p.u.

Circuit analysis using phasor/impedance, complex power calculation: A three-phase RLC circuit
is shown below. The reactance of the capacitor is 40 Ohm. The reactance of the inductor is 80 Ohm, and the
resistor is 8 Ohm. The sending end voltage is v₁ (t) = √2³45 cos(wt + 7) kV, where w = 27 · 60 rad/s. The
receiving end voltage is v2(t) = √√2³45 cos(wt) kV. The power base is 1000 MVA and the voltage base (L-L) is
345 kV.
Please compute total transferred complex power S₁ and S₂ in the physical unit.
Please draw a per-phase phasor/impedance diagram using per unit values.
Please compute the per unit complex power S₁ and S₂.
Find out the real power loss by the resistor, reactive power consumption by the inductor and the reactive
power generated by the series capacitor in per unit.
S₁
→>>>>
R
L
с
qoldingu
Figure
Vr

Expert Solution

Step 1: Summarize the given information.

For a three-phase RLC circuit, the following information is given:

base parameters: $S_{b} = 100 M V A,$ $V_{b} = 345 k V$
circuit parameter: $R = 8 Ω, X_{L} = 80 Ω, X_{C} = 40 Ω$
source voltage: $v_{1} (t) = \sqrt{2} \frac{345}{\sqrt{3}} c o s (ω t + \frac{π}{6}) k V, v_{2} (t) = \sqrt{2} \frac{345}{\sqrt{3}} c o s (ω t) k V .$

Step by step

Solved in 5 steps with 6 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Sinusoids and Phasors of Alternating Circuit

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.

Similar questions