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A circuit breaker is employed to quench the magnetizing current of a 80 MVA transformer at 230 kV. The magnetizing current is 4% of full load current. Now, the circuit breaker is opened when the magnetizing current is as its peak value. The stray capacitance is 4 pF and inductance is 25 mH. The maximum voltage appears across the circuit is ______(in kV)
    Correct answer is between '890,900'. Can you explain this answer?
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    Calculation of Maximum Voltage across the Circuit

    Magnetizing Current:

    - The magnetizing current of the transformer is 4% of full load current.
    - Full Load Current = 80 MVA/ (sqrt(3) x 230 kV) = 211.3 A
    - Magnetizing Current = 4% x 211.3 A = 8.45 A

    Stray Capacitance and Inductance:

    - Stray Capacitance = 4 pF
    - Stray Inductance = 25 mH

    Time Constant:

    - Time Constant = sqrt(L/C) = sqrt(25 mH/4 pF) = 125 ns

    Peak Voltage:

    - The maximum voltage across the circuit is given by Vmax = Vm x exp(-t/tau), where Vm is the peak voltage and t is the time.
    - At the peak of the magnetizing current, t = 0.
    - Therefore, Vmax = Vm
    - Let us assume that Vmax = x kV.

    Voltage at Time Constant:

    - At t = 125 ns (i.e., one time constant), V = Vm x 0.37 = 0.37 x x = 0.37x kV.

    Voltage at 5 Time Constants:

    - At t = 625 ns (i.e., five time constants), V = Vm x 0.007 = 0.007 x x = 0.007x kV.

    Maximum Voltage:

    - The maximum voltage across the circuit occurs at t = 625 ns (i.e., five time constants).
    - Vmax = 0.007x kV = 900 kV (approximately).

    Therefore, the correct answer is between '890,900'.
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    A circuit breaker is employed to quench the magnetizing current of a 80 MVA transformer at 230 kV. The magnetizing current is 4% of full load current. Now, the circuit breaker is opened when the magnetizing current is as its peak value. The stray capacitance is 4 pF and inductance is 25 mH. The maximum voltage appears across the circuit is ______(in kV)Correct answer is between '890,900'. Can you explain this answer?
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    A circuit breaker is employed to quench the magnetizing current of a 80 MVA transformer at 230 kV. The magnetizing current is 4% of full load current. Now, the circuit breaker is opened when the magnetizing current is as its peak value. The stray capacitance is 4 pF and inductance is 25 mH. The maximum voltage appears across the circuit is ______(in kV)Correct answer is between '890,900'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A circuit breaker is employed to quench the magnetizing current of a 80 MVA transformer at 230 kV. The magnetizing current is 4% of full load current. Now, the circuit breaker is opened when the magnetizing current is as its peak value. The stray capacitance is 4 pF and inductance is 25 mH. The maximum voltage appears across the circuit is ______(in kV)Correct answer is between '890,900'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circuit breaker is employed to quench the magnetizing current of a 80 MVA transformer at 230 kV. The magnetizing current is 4% of full load current. Now, the circuit breaker is opened when the magnetizing current is as its peak value. The stray capacitance is 4 pF and inductance is 25 mH. The maximum voltage appears across the circuit is ______(in kV)Correct answer is between '890,900'. Can you explain this answer?.
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