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In a 220 kV system, the inductance and capacitance up the circuit breaker location are 25 mH and 0.025mF respectively. The value of resistor required to be connected across the breaker contacts which will give no transient oscillations, is
- a)25 W
- b)250W
- c)500W
- d)1000W
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In a 220 kV system, the inductance and capacitance up the circuit brea...
The value of resistor required to be connected across the breaker contacts which will give no transient oscillations R is
R = 0.5* √(L/C)
R = 0.5 * √(25*10-3/(0.025*10-6))
R = 500 Ω
R = 0.5* √(L/C)
R = 0.5 * √(25*10-3/(0.025*10-6))
R = 500 Ω
Most Upvoted Answer
In a 220 kV system, the inductance and capacitance up the circuit brea...
Given:
- System voltage (V) = 220 kV = 220,000 V
- Inductance (L) = 25 mH = 0.025 H
- Capacitance (C) = 0.025 mF = 0.025 x 10^-6 F
To Find:
- Value of resistor (R) required to be connected across the breaker contacts
Solution:
Transient oscillations occur in a circuit due to the presence of inductance and capacitance. To eliminate these oscillations, a resistor needs to be connected across the breaker contacts. This resistor is known as a damping resistor.
The value of the damping resistor can be calculated using the formula:
R = 2√(L/C)
Calculation:
R = 2√(L/C)
= 2√(0.025 / 0.025 x 10^-6)
= 2√(0.025 x 10^6)
= 2√(25 x 10^3)
= 2 x 5 x 10^3
= 10 x 10^3
= 10,000 Ω
= 10 kΩ
The value of the resistor required to be connected across the breaker contacts to eliminate transient oscillations is 10 kΩ.
Answer:
The correct option is 500 Ω.
- System voltage (V) = 220 kV = 220,000 V
- Inductance (L) = 25 mH = 0.025 H
- Capacitance (C) = 0.025 mF = 0.025 x 10^-6 F
To Find:
- Value of resistor (R) required to be connected across the breaker contacts
Solution:
Transient oscillations occur in a circuit due to the presence of inductance and capacitance. To eliminate these oscillations, a resistor needs to be connected across the breaker contacts. This resistor is known as a damping resistor.
The value of the damping resistor can be calculated using the formula:
R = 2√(L/C)
Calculation:
R = 2√(L/C)
= 2√(0.025 / 0.025 x 10^-6)
= 2√(0.025 x 10^6)
= 2√(25 x 10^3)
= 2 x 5 x 10^3
= 10 x 10^3
= 10,000 Ω
= 10 kΩ
The value of the resistor required to be connected across the breaker contacts to eliminate transient oscillations is 10 kΩ.
Answer:
The correct option is 500 Ω.
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In a 220 kV system, the inductance and capacitance up the circuit breaker location are 25 mH and 0.025mF respectively. The value of resistor required to be connected across the breaker contacts which will give no transient oscillations, isa)25 Wb)250Wc)500Wd)1000WCorrect answer is option 'C'. Can you explain this answer?
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