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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)
Correct answer is '0.004'. Can you explain this answer?
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It is required to operate 150 A SCR in parallel with 250 A SCR with th...
Order to calculate the required resistance to be inserted in series with each SCR, we need to consider the current ratings and ON-State voltage drops of each SCR.
Let's assume the resistance to be inserted in series with the 150 A SCR is R1, and the resistance to be inserted in series with the 250 A SCR is R2.
We can use the formula for calculating current sharing in parallel resistances:
I1/I2 = R2/R1
Given that the total load is 400 A, the current through the 150 A SCR can be calculated as follows:
I1 = (150 A / (150 A + 250 A)) * 400 A
I1 = (150 A / 400 A) * 400 A
I1 = 150 A
Similarly, the current through the 250 A SCR can be calculated as follows:
I2 = (250 A / (150 A + 250 A)) * 400 A
I2 = (250 A / 400 A) * 400 A
I2 = 250 A
Next, we can calculate the voltage drops across each SCR:
V1 = 150 A * 1.2 V
V1 = 180 V
V2 = 250 A * 0.8 V
V2 = 200 V
Now, we can use the formula mentioned above to calculate the values of R1 and R2:
150 A / 250 A = R2 / R1
Simplifying the equation:
0.6 = R2 / R1
We know that V = I * R, so we can express the resistance values in terms of voltage drops:
R1 = V1 / I1
R1 = 180 V / 150 A
R1 = 1.2 Ω
R2 = V2 / I2
R2 = 200 V / 250 A
R2 = 0.8 Ω
Therefore, the value of resistance to be inserted in series with the 150 A SCR is 1.2 Ω, and the value of resistance to be inserted in series with the 250 A SCR is 0.8 Ω.
Let's assume the resistance to be inserted in series with the 150 A SCR is R1, and the resistance to be inserted in series with the 250 A SCR is R2.
We can use the formula for calculating current sharing in parallel resistances:
I1/I2 = R2/R1
Given that the total load is 400 A, the current through the 150 A SCR can be calculated as follows:
I1 = (150 A / (150 A + 250 A)) * 400 A
I1 = (150 A / 400 A) * 400 A
I1 = 150 A
Similarly, the current through the 250 A SCR can be calculated as follows:
I2 = (250 A / (150 A + 250 A)) * 400 A
I2 = (250 A / 400 A) * 400 A
I2 = 250 A
Next, we can calculate the voltage drops across each SCR:
V1 = 150 A * 1.2 V
V1 = 180 V
V2 = 250 A * 0.8 V
V2 = 200 V
Now, we can use the formula mentioned above to calculate the values of R1 and R2:
150 A / 250 A = R2 / R1
Simplifying the equation:
0.6 = R2 / R1
We know that V = I * R, so we can express the resistance values in terms of voltage drops:
R1 = V1 / I1
R1 = 180 V / 150 A
R1 = 1.2 Ω
R2 = V2 / I2
R2 = 200 V / 250 A
R2 = 0.8 Ω
Therefore, the value of resistance to be inserted in series with the 150 A SCR is 1.2 Ω, and the value of resistance to be inserted in series with the 250 A SCR is 0.8 Ω.
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Community Answer
It is required to operate 150 A SCR in parallel with 250 A SCR with th...
Dynamic resistance of 150 A. SCR1 = 1.2/150 = 8 mΩ
Dynamic resistance of 250 A SCR 2 = 0.8/250
= 3.2 mΩ
= 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
Current shared by SCR1 =
Current shared by SCR2 =
⇒ 16 + 5 Rs = 3 Rs + 24
⇒ Rs = 0.004
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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)Correct answer is '0.004'. Can you explain this answer?
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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)Correct answer is '0.004'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 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)Correct answer is '0.004'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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)Correct answer is '0.004'. Can you explain this answer?.
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)Correct answer is '0.004'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 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)Correct answer is '0.004'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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)Correct answer is '0.004'. Can you explain this answer?.
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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)Correct answer is '0.004'. Can you explain this answer?, a detailed solution for 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)Correct answer is '0.004'. Can you explain this answer? has been provided alongside types of 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)Correct answer is '0.004'. Can you explain this answer? theory, EduRev gives you an
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