Practice Questions: Parallel Operation of Thyristors

# Practice Questions: Parallel Operation of Thyristors

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## 5 Questions MCQ Test Power Electronics | Practice Questions: Parallel Operation of Thyristors

Practice Questions: Parallel Operation of Thyristors for Electrical Engineering (EE) 2023 is part of Power Electronics preparation. The Practice Questions: Parallel Operation of Thyristors questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Practice Questions: Parallel Operation of Thyristors MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Questions: Parallel Operation of Thyristors below.
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Practice Questions: Parallel Operation of Thyristors - Question 1

### SCR with a rating of 1200 V and 250 A are available to be used in a string to handle 8 kV and 1kA. The number of parallel and series units required are respectively _____ (Take derating factor is 0.2)

Detailed Solution for Practice Questions: Parallel Operation of Thyristors - Question 1

Derating factor = 1 – string efficiency *Answer can only contain numeric values
Practice Questions: Parallel Operation of Thyristors - Question 2

### The figure below shows two thyristors each rated 500 A (continuous) sharing a load current. current through thyristor T2 is 120 A the current through T1 will be nearly equal to ___ (in A) Detailed Solution for Practice Questions: Parallel Operation of Thyristors - Question 2

Concept

Matched-pair SCR’s are generally available for parallel connection, but they are very expensive. With unmatched SCR’s equal current sharing is enforced by adding a low-value resistor or inductor in series with each SCR. Forced current sharing using equal-value resistors is shown in Figure 2. The basic requirement is to make current I1 close to l2, a maximum difference of 20% is accept across SCR2, the value of R can be obtained from. I1R + V1 = I2R + V2

R = (V1 – V2)/(I2 – I1)

Calculation:

Voltage across 0.05 ohm resistor  = 120 × 0.05 = 6 V
⇒ I2 = 6/0.06 = 100 A
Current through T1 will be nearly equal to 100  A

*Answer can only contain numeric values
Practice Questions: Parallel Operation of Thyristors - Question 3

### A 300 A thyristor is to be operated in parallel with a 400 A thyristor their on state voltage drops are 1.8 V and 1 V respectively. The resistance to be connected in series with each thyristor so that the current through the combination is 700 A and each of them is fully loaded is ___________(in mΩ)

Detailed Solution for Practice Questions: Parallel Operation of Thyristors - Question 3

Concept

Matched-pair SCR’s are generally available for parallel connection, but they are very expensive. With unmatched SCR’s equal current sharing is enforced by adding a low-value resistor or inductor in series with each SCR. Forced current sharing using equal-value resistors is shown in Figure 2. The basic requirement is to make current I1 close to l2, a maximum difference of 20% is accept across SCR2, the value of R can be obtained from. I1R + V1 = I2R + V2

R = (V1 – V2)/(I2 – I1)

Calculation:

In parallel anode to cathode voltage drops are same

1.8 + 300 R = 1.0 + 400 R

⇒ 100 R = 0.8

⇒ R = 0.008 = 8 mΩ

*Answer can only contain numeric values
Practice Questions: Parallel Operation of Thyristors - Question 4

A 200 A thyristor is to be operated in parallel with a 300 A thyristor their on state voltage drops are 1.5 V and 1.2 V respectively. What is the value of resistance to be connected in series with each thyristor so that the current through the combination is 500 A and each of them is fully loaded. ____ (in Ω)

Detailed Solution for Practice Questions: Parallel Operation of Thyristors - Question 4

Concept

Matched-pair SCR’s are generally available for parallel connection, but they are very expensive. With unmatched SCR’s equal current sharing is enforced by adding a low-value resistor or inductor in series with each SCR. Forced current sharing using equal-value resistors is shown in Figure 2. The basic requirement is to make current I1 close to l2, a maximum difference of 20% is accept across SCR2, the value of R can be obtained from. I1R + V1 = I2R + V2

R = (V1 – V2)/(I2 – I1)

In parallel anode to cathode voltage drops are same

1.5 + 200 R = 1.2 + 300 R

⇒ R = 0.003 Ω

*Answer can only contain numeric values
Practice Questions: Parallel Operation of Thyristors - Question 5

It is required to operate 150 A SCR in parallel with 250 A SCR with their respective ON-State voltage drop of 1.2 V and 0.8 V. Calculate the value of resistance to be inserted in series with each SCR so that they share the total load of 400 A in proportion to their current ratings.___(in Ω)

Detailed Solution for Practice Questions: Parallel Operation of Thyristors - Question 5

Dynamic resistance of 150 A. SCR1 = 1.2/150 = 8 mΩ

Dynamic resistance of 250 A SCR 2 = 0.8/250
= 3.2 mΩ

Let RS be the resistance inserted in series with each SCR with this
Current shared by SCR1 = Current shared by SCR2 =  ⇒ 16 + 5 Rs = 3 Rs + 24
⇒ Rs = 0.004

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