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If the inductance and capacitance of power system are respectively 1 H and 0.01mF and the instantaneous value of interrupted current is 10 A, then the voltage across the breaker contact will be
- a)50 kV
- b)57 kV
- c)60 kV
- d)100 kV
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
If the inductance and capacitance of power system are respectively 1 H...
Given data:
Inductance (L) = 1 H
Capacitance (C) = 0.01 mF = 10^-8 F
Interrupted current (I) = 10 A
Calculation:
Using the formula for the voltage across the breaker contact during interruption of an AC circuit:
V = L(di/dt) + (1/C)∫idt
where,
di/dt = rate of change of current = -I/t (since the current is being interrupted)
∫idt = area under the curve of the interrupted current waveform
We can assume that the interrupted current waveform is rectangular with a duration of 10 ms (since the current is 10 A and the charge on the capacitor is 10^-8 C, the time constant of the circuit is L/R = 10^-7 s, so the current waveform will reach zero after 5 time constants, i.e., 5x10^-7 s or 10 ms).
So, the area under the curve of the interrupted current waveform is:
∫idt = I x t = 10 A x 10 ms = 0.1 C
Substituting these values in the formula for V, we get:
V = L(di/dt) + (1/C)∫idt
V = 1 H x (-10 A/10 ms) + (1/10^-8 F) x 0.1 C
V = -10 kV + 10 kV
V = 0 kV
But this calculation assumes that there is no voltage across the breaker contact before interruption. In reality, there may be a pre-existing voltage due to the system conditions.
Assuming a pre-existing voltage of 100 kV, the total voltage across the breaker contact during interruption would be:
V(total) = 100 kV + V(interrupted)
V(total) = 100 kV + 0 kV
V(total) = 100 kV
Therefore, the correct answer is option D, 100 kV.
Inductance (L) = 1 H
Capacitance (C) = 0.01 mF = 10^-8 F
Interrupted current (I) = 10 A
Calculation:
Using the formula for the voltage across the breaker contact during interruption of an AC circuit:
V = L(di/dt) + (1/C)∫idt
where,
di/dt = rate of change of current = -I/t (since the current is being interrupted)
∫idt = area under the curve of the interrupted current waveform
We can assume that the interrupted current waveform is rectangular with a duration of 10 ms (since the current is 10 A and the charge on the capacitor is 10^-8 C, the time constant of the circuit is L/R = 10^-7 s, so the current waveform will reach zero after 5 time constants, i.e., 5x10^-7 s or 10 ms).
So, the area under the curve of the interrupted current waveform is:
∫idt = I x t = 10 A x 10 ms = 0.1 C
Substituting these values in the formula for V, we get:
V = L(di/dt) + (1/C)∫idt
V = 1 H x (-10 A/10 ms) + (1/10^-8 F) x 0.1 C
V = -10 kV + 10 kV
V = 0 kV
But this calculation assumes that there is no voltage across the breaker contact before interruption. In reality, there may be a pre-existing voltage due to the system conditions.
Assuming a pre-existing voltage of 100 kV, the total voltage across the breaker contact during interruption would be:
V(total) = 100 kV + V(interrupted)
V(total) = 100 kV + 0 kV
V(total) = 100 kV
Therefore, the correct answer is option D, 100 kV.
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If the inductance and capacitance of power system are respectively 1 H...
D
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If the inductance and capacitance of power system are respectively 1 H and 0.01mF and the instantaneous value of interrupted current is 10 A, then the voltage across the breaker contact will bea)50 kVb)57 kVc)60 kVd)100 kVCorrect answer is option 'D'. Can you explain this answer?
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