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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Test: Interconnecting Thyristors - Electrical Engineering (EE) MCQ

Test: Interconnecting Thyristors - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test - Test: Interconnecting Thyristors

Test: Interconnecting Thyristors for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Interconnecting Thyristors questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Interconnecting Thyristors MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Interconnecting Thyristors below.
Solutions of Test: Interconnecting Thyristors questions in English are available as part of our course for Electrical Engineering (EE) & Test: Interconnecting Thyristors solutions in Hindi for Electrical Engineering (EE) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Interconnecting Thyristors | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Interconnecting Thyristors - Question 1
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SCRs are connected in parallel to fulfill the ___________ demand

Detailed Solution for Test: Interconnecting Thyristors - Question 1

Number of devices connected in parallel can carry huge amounts of current.

Test: Interconnecting Thyristors - Question 2
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The term used to measure the degree of utilization of SCRs connected in series & parallel is

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String Efficiency = Rating of the whole string/(rating of one SCR x number of SCRs)

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Test: Interconnecting Thyristors - Question 3
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To have maximum possible string efficiency

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Having similar ratings does not mean they have similar charc.

Test: Interconnecting Thyristors - Question 4
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For a string voltage of 3300 V, let there be six series connected SCRs each of voltage 600V. Then the string efficiency is
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 String efficiency = 3300/(6 x 600) = 98.54.

Test: Interconnecting Thyristors - Question 5
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The measure of reliability of string is given by the factor
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 DRF is de-rating factor given by the above expression.

Test: Interconnecting Thyristors - Question 6
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When an extra SCR is connected in series with a string
Detailed Solution for Test: Interconnecting Thyristors - Question 6

DRF = 1 – String efficiency
String Efficiency = Rating of the whole string/(rating of one SCR x number of SCRs)
Extra SCR will reduce the string efficiency which in turn increase the DRF.

Test: Interconnecting Thyristors - Question 7
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The most practical way of obtaining a uniform distribution of series connected SCRs is to
Detailed Solution for Test: Interconnecting Thyristors - Question 7

For uniform distribution of voltage across series connected SCRs, a resistor of value R in parallel with each series connected SCR.

Test: Interconnecting Thyristors - Question 8
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3 SCRs are connected in series. The string efficiency is 91%. SCRs 1, 2 & 3 have leakage currents 4 mA, 8 mA & 12 mA. Which SCR will block more voltage?
Detailed Solution for Test: Interconnecting Thyristors - Question 8

The SCR with lower leakage current block more voltage.

Test: Interconnecting Thyristors - Question 9
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Two parallel connect SCRs have same voltage drop (Vt) having rated current = 2I1. SCR-1 carries a current of I1=2.6 A whereas SCR-2 carries a current of I2=1.4 A. Find the string efficiency.
Detailed Solution for Test: Interconnecting Thyristors - Question 9

The total current would be I1+I2 & rated current is 2I1
String efficiency = (I1+I2)/2I1.

Test: Interconnecting Thyristors - Question 10
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SCRs with a rating of 1000 V & 200 A are available to be used in a string to handle 6 KV & 1 KV. Calculate the number of series & parallel units required in case the de-rating factor is 0.1. (Round off the fraction to the greatest & nearest integer)
Detailed Solution for Test: Interconnecting Thyristors - Question 10

 DRF = 1-S.E
Therefore
0.1 = (1-6000/1000Ns) = (1-1000/200Np)
Ns = 6.6 = 7(say)
Np = 5.5 = 6(say)

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