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Test: Heating, cooling and mounting of thyristors - Electrical Engineering (EE) MCQ


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5 Questions MCQ Test - Test: Heating, cooling and mounting of thyristors

Test: Heating, cooling and mounting of thyristors for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Heating, cooling and mounting of thyristors questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Heating, cooling and mounting of thyristors MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Heating, cooling and mounting of thyristors below.
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Test: Heating, cooling and mounting of thyristors - Question 1

National Highways Authority of India (NHAI) has inked initial pacts with how many technical institutes to encourage them for voluntary adoption of highway stretches for research etc?

Detailed Solution for Test: Heating, cooling and mounting of thyristors - Question 1

The correct answer is 200.

  • National Highways Authority of India (NHAI) has inked initial pacts with 200 technical institutes to encourage them for voluntary adoption of highway stretches for research etc.
  • So far, 18 IITs, 27 NITs and 207 other reputed engineering colleges have consented for the MoU.
  • Out of over 250 institutes which have consented for the MoU, 200 have become signatories.
*Answer can only contain numeric values
Test: Heating, cooling and mounting of thyristors - Question 2

An SCR has ambient temperature of 25°C and Junction temperature of 50°C and Average power dissipated of 25 watt. If the junction temperature is reduced to 35°C, the percentage reduction in thermal resistance _______ 


Detailed Solution for Test: Heating, cooling and mounting of thyristors - Question 2

Concept

Here,

Pav = Average rate of heat generated at a thyristor Junction

QJC = thermal resistance between the junction temperature TJ and case temperature TC

QCS = thermal resistance between the case temperature TC and sink temperature

QSA = thermal resistance between the sink temperature TS and ambient temperature TA

 
Where QJA = QJC + QCS + QSA

Calculations:

TA = 25°C

IJ = 50°C

Pav = 25 watt


 ∴ % change in thermal resistance

 
 

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Test: Heating, cooling and mounting of thyristors - Question 3

The thermal resistance for the thyristor – sin k combination are QJC = 0.18 and ACS = 0.09 °C/W for a heat sin k temperature of 75°C and maximum junction temperature is 130°C . In case the heat sin k temperature is brought down 70°C by forced coaling find the percentage increase in the device rating.

Detailed Solution for Test: Heating, cooling and mounting of thyristors - Question 3


Thus total average power loss in the thyristor – sink combination is 203.70 W. with improved coaling

Thyristor rating is proportional to the square root of average power loss

∴ Percentage Increase in thyristor rating

*Answer can only contain numeric values
Test: Heating, cooling and mounting of thyristors - Question 4

For a thysistor maximum junction temperature is 200°C. The thermal resistance for the thyristor sink combination are θJC = 0.15 and θCS = 0.05c/w. Initially the heat sink temperature of 80° C and it is brought down to 70° C by force cooling, the percentage increase in the device rating is – (in%)


Detailed Solution for Test: Heating, cooling and mounting of thyristors - Question 4


From the above equivalent circuit

TJ = Ts + PavJC + θCS)

Thyristor rating is proportional to the square root of average power loss

% increase in thyristor rating


 x 100 = 4.08%

Test: Heating, cooling and mounting of thyristors - Question 5

The usual way to accomplish higher gate current for improved di/dt rating is by using

Detailed Solution for Test: Heating, cooling and mounting of thyristors - Question 5

Pilot SCR is an SCR which is fired which activates the firing circuit and fires the main SCR.

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