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In a 132 kV system, the phase to ground capacitanc
e is 0.01 µF and the inductance is 4H. Calculate the voltage appearing across the electrodes of C.B. If a magnetizing current at 5 amp is interrupted. Calculate the critical resistance to be connected, to eliminate restriking.
a)100 kV, 10 kohm
b)10 kV, 100 kohm
c)100 V, 10000ohm
d)100 kV, 100 kohm
Correct answer is option 'A'. Can you explain this answer?
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In a 132 kV system, the phase to ground capacitanc... moree is 0.01 µF...
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In a 132 kV system, the phase to ground capacitanc... moree is 0.01 µF...
The given question asks us to calculate the voltage appearing across the electrodes of a circuit breaker (C.B.) when a magnetizing current of 5 A is interrupted in a 132 kV system. We are also required to determine the critical resistance that needs to be connected to eliminate restriking.
Given data:
Phase to ground capacitance (C) = 0.01 µF
Inductance (L) = 4 H
Magnetizing current (I) = 5 A
System voltage (V) = 132 kV
To calculate the voltage appearing across the electrodes of the circuit breaker, we can use the formula:
V = √(2 * π * f * L * I)
Where:
V = Voltage appearing across the electrodes
f = Frequency of the system
Now, let's find the frequency of the system:
The frequency (f) can be calculated using the formula:
f = 1 / (2 * π * √(L * C))
Substituting the given values:
f = 1 / (2 * π * √(4 * 0.01 * 10^-6))
Simplifying the equation:
f = 1 / (2 * π * √(4 * 10^-8))
f = 1 / (2 * π * 2 * 10^-4)
f = 1 / (4 * π * 10^-4)
f = 1 / (4 * 3.14 * 10^-4)
f = 1 / (12.56 * 10^-4)
f = 1 / 0.01256
f ≈ 79.52 Hz
Now, substituting the values into the voltage formula:
V = √(2 * π * f * L * I)
V = √(2 * π * 79.52 * 4 * 5)
Simplifying the equation:
V = √(2 * π * 79.52 * 4 * 5)
V = √(2 * 3.14 * 79.52 * 4 * 5)
V = √(100 * 3.14 * 79.52)
V = √(314 * 79.52)
V = √(24944.28)
V ≈ 157.84 kV
Therefore, the voltage appearing across the electrodes of the circuit breaker is approximately 157.84 kV.
To eliminate restriking, a critical resistance needs to be connected. The critical resistance can be calculated using the formula:
R = V^2 / (2 * π * f * C)
Substituting the given values:
R = (157.84 * 10^3)^2 / (2 * π * 79.52 * 0.01 * 10^-6)
Simplifying the equation:
R = (157.84 * 10^3)^2 / (2 * 3.14 * 79.52 * 0.01 * 10^-6)
R = (24944.28 * 10^6) / (6.28 * 0.7952 * 10^-2)
R = (24944.28 * 10^6) / (4.9936 * 10^-2)
R = 498.88 * 10
Given data:
Phase to ground capacitance (C) = 0.01 µF
Inductance (L) = 4 H
Magnetizing current (I) = 5 A
System voltage (V) = 132 kV
To calculate the voltage appearing across the electrodes of the circuit breaker, we can use the formula:
V = √(2 * π * f * L * I)
Where:
V = Voltage appearing across the electrodes
f = Frequency of the system
Now, let's find the frequency of the system:
The frequency (f) can be calculated using the formula:
f = 1 / (2 * π * √(L * C))
Substituting the given values:
f = 1 / (2 * π * √(4 * 0.01 * 10^-6))
Simplifying the equation:
f = 1 / (2 * π * √(4 * 10^-8))
f = 1 / (2 * π * 2 * 10^-4)
f = 1 / (4 * π * 10^-4)
f = 1 / (4 * 3.14 * 10^-4)
f = 1 / (12.56 * 10^-4)
f = 1 / 0.01256
f ≈ 79.52 Hz
Now, substituting the values into the voltage formula:
V = √(2 * π * f * L * I)
V = √(2 * π * 79.52 * 4 * 5)
Simplifying the equation:
V = √(2 * π * 79.52 * 4 * 5)
V = √(2 * 3.14 * 79.52 * 4 * 5)
V = √(100 * 3.14 * 79.52)
V = √(314 * 79.52)
V = √(24944.28)
V ≈ 157.84 kV
Therefore, the voltage appearing across the electrodes of the circuit breaker is approximately 157.84 kV.
To eliminate restriking, a critical resistance needs to be connected. The critical resistance can be calculated using the formula:
R = V^2 / (2 * π * f * C)
Substituting the given values:
R = (157.84 * 10^3)^2 / (2 * π * 79.52 * 0.01 * 10^-6)
Simplifying the equation:
R = (157.84 * 10^3)^2 / (2 * 3.14 * 79.52 * 0.01 * 10^-6)
R = (24944.28 * 10^6) / (6.28 * 0.7952 * 10^-2)
R = (24944.28 * 10^6) / (4.9936 * 10^-2)
R = 498.88 * 10
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In a 132 kV system, the phase to ground capacitanc... moree is 0.01 µF...
In a 132 kilo old system the line to ground capacitance is 0.01 megapired and the inductance is 4 h what is value of voltage apparent cross the CB pole if the measuring current to be interepted is 5 AMS
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In a 132 kV system, the phase to ground capacitanc... moree is 0.01 µF and the inductance is 4H. Calculate the voltage appearing across the electrodes of C.B. If a magnetizing current at 5 amp is interrupted. Calculate the critical resistance to be connected, to eliminate restriking.a)100 kV, 10 kohmb)10 kV, 100 kohmc)100 V, 10000ohmd)100 kV, 100 kohmCorrect answer is option 'A'. Can you explain this answer?
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