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In a 220 kV system, the inductance and capacitance up to the circuit breaker location are 25 mH and 0.025 μF respectively. The value of resistor required to be connected across the breaker contacts which will give no transient oscillations, is
  • a)
    25 Ω
  • b)
    250 Ω
  • c)
    500 Ω
  • d)
    1000 Ω
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
In a 220 kV system, the inductance and capacitance up to the circuit b...
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
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Most Upvoted Answer
In a 220 kV system, the inductance and capacitance up to the circuit b...
Solution:

Given data:

Inductance (L) = 25 mH = 0.025 H

Capacitance (C) = 0.025 F

Voltage (V) = 220 kV

Resistor required to be connected across the breaker contacts (R) = ?

To prevent the transient oscillations,

R = √(L/C) = √(0.025/0.025) = 1 Ω

Therefore, the value of resistor required to be connected across the breaker contacts which will give no transient oscillations is 1 Ω or 1000 Ω (Option C).

Explanation:

When a circuit breaker opens, an arc is generated across the contacts due to the interruption of the current. This arc creates a transient oscillation in the circuit due to the inductance and capacitance of the system. To prevent this oscillation, a resistor is connected across the contacts of the circuit breaker.

The value of the resistor required to be connected across the contacts can be calculated using the formula:

R = √(L/C)

where L is the inductance of the system and C is the capacitance of the system.

In this case, the inductance (L) is 25 mH or 0.025 H and the capacitance (C) is 0.025 F. Substituting these values in the above formula, we get:

R = √(0.025/0.025) = 1 Ω

Therefore, the value of resistor required to be connected across the breaker contacts which will give no transient oscillations is 1 Ω or 1000 Ω (Option C).
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