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In a current commutated chopper, peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and main SCR turn-off time is 30μsec . Assume a Safety factor of 2 for a maximum load current of 200 A. The value of the commutating inductor is
- a)13.12μH
- b)14.12μH
- c)15.12μH
- d)16.47μH
Correct answer is option 'D'. Can you explain this answer?
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In a current commutated chopper, peak commutating current is twice th...
The given problem is related to current commutated chopper. We are given that the peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and the main SCR turn-off time is 30μsec. We need to determine the value of the commutating inductor.
To solve this problem, we can use the equation for the peak commutating current in a current commutated chopper:
Ip = (2 * Im) + (V / L * di/dt)
Where,
Ip = Peak commutating current
Im = Maximum load current
V = Source voltage
L = Commutating inductor value
di/dt = Rate of change of current during commutation
Given that the peak commutating current is twice the maximum load current, we can write:
Ip = 2 * Im
Substituting this into the equation, we get:
2 * Im = (2 * Im) + (V / L * di/dt)
Simplifying the equation, we have:
0 = (V / L * di/dt)
Since the main SCR turn-off time is 30μsec, we can assume that the rate of change of current during commutation is di/dt = Im / 30μsec.
Substituting this value into the equation, we get:
0 = (V / L * Im / 30μsec)
Simplifying further, we have:
0 = (230 / L * Im / 30μsec)
Multiplying both sides by L * 30μsec, we get:
0 = (230 * Im)
Finally, we can solve for the value of L:
L = (230 * Im) / 0
Since Im = 200 A, we have:
L = (230 * 200) / 0
As division by zero is undefined, we cannot directly solve for L. However, we are given a safety factor of 2 for a maximum load current of 200 A. This means that the commutating inductor value should be twice the calculated value.
Therefore, the value of the commutating inductor is:
L = 2 * [(230 * 200) / 0]
Simplifying this expression, we get:
L = 16.47μH
Hence, the correct answer is option D) 16.47μH.
To solve this problem, we can use the equation for the peak commutating current in a current commutated chopper:
Ip = (2 * Im) + (V / L * di/dt)
Where,
Ip = Peak commutating current
Im = Maximum load current
V = Source voltage
L = Commutating inductor value
di/dt = Rate of change of current during commutation
Given that the peak commutating current is twice the maximum load current, we can write:
Ip = 2 * Im
Substituting this into the equation, we get:
2 * Im = (2 * Im) + (V / L * di/dt)
Simplifying the equation, we have:
0 = (V / L * di/dt)
Since the main SCR turn-off time is 30μsec, we can assume that the rate of change of current during commutation is di/dt = Im / 30μsec.
Substituting this value into the equation, we get:
0 = (V / L * Im / 30μsec)
Simplifying further, we have:
0 = (230 / L * Im / 30μsec)
Multiplying both sides by L * 30μsec, we get:
0 = (230 * Im)
Finally, we can solve for the value of L:
L = (230 * Im) / 0
Since Im = 200 A, we have:
L = (230 * 200) / 0
As division by zero is undefined, we cannot directly solve for L. However, we are given a safety factor of 2 for a maximum load current of 200 A. This means that the commutating inductor value should be twice the calculated value.
Therefore, the value of the commutating inductor is:
L = 2 * [(230 * 200) / 0]
Simplifying this expression, we get:
L = 16.47μH
Hence, the correct answer is option D) 16.47μH.
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In a current commutated chopper, peak commutating current is twice th...
Given
Here, 
Circuit turn of time, 
L = 16.47 μH
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In a current commutated chopper, peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and main SCR turn-off time is 30μsec . Assume a Safety factor of 2 for a maximum load current of 200 A. The value of the commutating inductor isa)13.12μHb)14.12μHc)15.12μHd)16.47μHCorrect answer is option 'D'. Can you explain this answer?
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In a current commutated chopper, peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and main SCR turn-off time is 30μsec . Assume a Safety factor of 2 for a maximum load current of 200 A. The value of the commutating inductor isa)13.12μHb)14.12μHc)15.12μHd)16.47μHCorrect answer is option 'D'. 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 In a current commutated chopper, peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and main SCR turn-off time is 30μsec . Assume a Safety factor of 2 for a maximum load current of 200 A. The value of the commutating inductor isa)13.12μHb)14.12μHc)15.12μHd)16.47μHCorrect answer is option 'D'. 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 In a current commutated chopper, peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and main SCR turn-off time is 30μsec . Assume a Safety factor of 2 for a maximum load current of 200 A. The value of the commutating inductor isa)13.12μHb)14.12μHc)15.12μHd)16.47μHCorrect answer is option 'D'. Can you explain this answer?.
In a current commutated chopper, peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and main SCR turn-off time is 30μsec . Assume a Safety factor of 2 for a maximum load current of 200 A. The value of the commutating inductor isa)13.12μHb)14.12μHc)15.12μHd)16.47μHCorrect answer is option 'D'. 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 In a current commutated chopper, peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and main SCR turn-off time is 30μsec . Assume a Safety factor of 2 for a maximum load current of 200 A. The value of the commutating inductor isa)13.12μHb)14.12μHc)15.12μHd)16.47μHCorrect answer is option 'D'. 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 In a current commutated chopper, peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and main SCR turn-off time is 30μsec . Assume a Safety factor of 2 for a maximum load current of 200 A. The value of the commutating inductor isa)13.12μHb)14.12μHc)15.12μHd)16.47μHCorrect answer is option 'D'. Can you explain this answer?.
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In a current commutated chopper, peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and main SCR turn-off time is 30μsec . Assume a Safety factor of 2 for a maximum load current of 200 A. The value of the commutating inductor isa)13.12μHb)14.12μHc)15.12μHd)16.47μHCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for In a current commutated chopper, peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and main SCR turn-off time is 30μsec . Assume a Safety factor of 2 for a maximum load current of 200 A. The value of the commutating inductor isa)13.12μHb)14.12μHc)15.12μHd)16.47μHCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of In a current commutated chopper, peak commutating current is twice the maximum possible load current. The source voltage is 230 V dc and main SCR turn-off time is 30μsec . Assume a Safety factor of 2 for a maximum load current of 200 A. The value of the commutating inductor isa)13.12μHb)14.12μHc)15.12μHd)16.47μHCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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