The solution is largely correct but for a few errors.. Also note that the generator is represented by a voltage source and an impedance in series. Likewise the motor.please taking into consideration the following comment can you please resolve the question = 20Kr x 20kv.20k| Kv base (new) =For Section 3Ky base (nero)=Ky base,(new)200kvFindingLow voltage rating xKV bosehighHigh Voltage rating voltage side= 20kr200 KV20KYX 2UD KYDnnuo per unit empedanceWehave to remember that odd MVA & Old KWAis taken for each element from griven rating inthe equation. That is for

The question is about per unit system.

Answered: = 20kr x 20kv. 20k Kv base (new) - For…

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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The solution is largely correct but for a few errors.. Also note that the generator is represented by a voltage source and an impedance in series. Likewise the motor.

please taking into consideration the following comment can you please resolve the question

= 20Kr x 20kv.
20k
| Kv base (new) =
For Section 3
Ky base (nero)
=
Ky base,
(new)
200kv
Finding
Low voltage rating x
KV bose
high
High Voltage rating voltage side
= 20kr
200 KV
20KY
X 2UD KY
Dn
nuo per unit empedance
We
have to remember that odd MVA & Old KWA
is taken for each element from griven rating in
the equation. That is for

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