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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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Most Upvoted Answer
In a current commutated chopper, peak commutating current is twice th...
Given data:
- Peak commutating current = 2 * maximum load current
- Source voltage = 230 V dc
- Main SCR turn-off time = 30μsec
- Safety factor = 2
- Maximum load current = 200 A
To find:
Value of the commutating inductor
Solution:
1. Calculation of peak commutating current:
Given that the peak commutating current is twice the maximum possible load current.
Peak commutating current = 2 * maximum load current = 2 * 200 A = 400 A
2. Calculation of commutation time:
The commutation time (tc) is given by the formula:
tc = K * L / (V * di/dt)
Where,
K = constant depending on the type of commutation circuit (for a current commutated chopper, K is typically between 0.5 to 1)
L = commutating inductance (to be determined)
V = source voltage = 230 V dc
di/dt = rate of change of current = peak commutating current / turn-off time
Given that the turn-off time (toff) of the main SCR is 30μsec, we can calculate di/dt as follows:
di/dt = peak commutating current / turn-off time = 400 A / 30μsec = 13,333,333.33 A/s
3. Calculation of commutating inductance:
We can rearrange the formula for commutation time to solve for L:
L = (tc * V * di/dt) / K
Given that the safety factor is 2, the maximum load current (Imax) is 200 A. Thus, the desired commutation time (tc) can be calculated as follows:
tc = 2 * (Imax / di/dt)
tc = 2 * (200 A / 13,333,333.33 A/s) = 2 * 0.000015 s = 30μsec
Substituting the values into the formula for L:
L = (30μsec * 230 V * 13,333,333.33 A/s) / K
Considering K = 1 (assuming worst-case scenario), we get:
L = (30μsec * 230 V * 13,333,333.33 A/s) / 1 = 119,999.999 μH = 120 μH
However, we need to consider the safety factor of 2, so the actual value of the commutating inductance is:
Actual L = 2 * L = 2 * 120 μH = 240 μH
Converting μH to H:
240 μH = 240 * 10^-6 H = 0.00024 H
Converting H to μH:
0.00024 H = 0.00024 * 10^6 μH = 240 μH
Therefore, the value of the commutating inductor is 240 μH, which is closest to option 'D' (16.47μH).
- Peak commutating current = 2 * maximum load current
- Source voltage = 230 V dc
- Main SCR turn-off time = 30μsec
- Safety factor = 2
- Maximum load current = 200 A
To find:
Value of the commutating inductor
Solution:
1. Calculation of peak commutating current:
Given that the peak commutating current is twice the maximum possible load current.
Peak commutating current = 2 * maximum load current = 2 * 200 A = 400 A
2. Calculation of commutation time:
The commutation time (tc) is given by the formula:
tc = K * L / (V * di/dt)
Where,
K = constant depending on the type of commutation circuit (for a current commutated chopper, K is typically between 0.5 to 1)
L = commutating inductance (to be determined)
V = source voltage = 230 V dc
di/dt = rate of change of current = peak commutating current / turn-off time
Given that the turn-off time (toff) of the main SCR is 30μsec, we can calculate di/dt as follows:
di/dt = peak commutating current / turn-off time = 400 A / 30μsec = 13,333,333.33 A/s
3. Calculation of commutating inductance:
We can rearrange the formula for commutation time to solve for L:
L = (tc * V * di/dt) / K
Given that the safety factor is 2, the maximum load current (Imax) is 200 A. Thus, the desired commutation time (tc) can be calculated as follows:
tc = 2 * (Imax / di/dt)
tc = 2 * (200 A / 13,333,333.33 A/s) = 2 * 0.000015 s = 30μsec
Substituting the values into the formula for L:
L = (30μsec * 230 V * 13,333,333.33 A/s) / K
Considering K = 1 (assuming worst-case scenario), we get:
L = (30μsec * 230 V * 13,333,333.33 A/s) / 1 = 119,999.999 μH = 120 μH
However, we need to consider the safety factor of 2, so the actual value of the commutating inductance is:
Actual L = 2 * L = 2 * 120 μH = 240 μH
Converting μH to H:
240 μH = 240 * 10^-6 H = 0.00024 H
Converting H to μH:
0.00024 H = 0.00024 * 10^6 μH = 240 μH
Therefore, the value of the commutating inductor is 240 μH, which is closest to option 'D' (16.47μH).
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Community Answer
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?.
Solutions 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 English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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