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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'.
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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