EBK POWER SYSTEM ANALYSIS AND DESIGN
EBK POWER SYSTEM ANALYSIS AND DESIGN
6th Edition
ISBN: 9781305886957
Author: Glover
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 10, Problem 10.1P
To determine

The value of the CT secondary current (I’) and VT secondary voltage (V’).

Expert Solution & Answer
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Answer to Problem 10.1P

  V'=115VI'=4.184A

Explanation of Solution

Given Information:

  VLL=345KVS3ϕ=600MVApf=0.95laggingn1CTRatio=1200:5nVTRatio=3000:1 .

Calculation:

The secondary voltage of the voltage transformer (V’) is,

  V'=1nVLL

Substitute the values in the above equation,

  V'=13000×345×103V'=115V

Now, the current (I) entering the primary of the current transformer is,

  I=S3ϕ3VLL

On substituting the values in the above equation,

And get,

  I=600×1063×(345×103)I=600×106597.55×103I=1004A

Here, the CT equivalent circuit is,

  EBK POWER SYSTEM ANALYSIS AND DESIGN, Chapter 10, Problem 10.1P

Fig: CT equivalent circuit diagram

Now, the secondary current of CT is,

  I2=1n1×I

Substituting the values in the above equation,

  I2=51200×1004I2=4.184A

From the above equivalent circuit diagram, the secondary current consists of two components, there are IeandI' . Therefore the expression of the secondary current of CT is,

  I2=I'+Ie

Whereas, I’ is the relay current and Ie is the exciting current.

CT error is Zero in the given question.

So that the exciting current is Zero, (Ie=0), therefore,

  I2=I'+Ie4.184A=I'+0I'=4.184A

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EBK POWER SYSTEM ANALYSIS AND DESIGN

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