Question

In a 220kV system, the inductance and capacitance up to the circuit breaker location are 25 mH and 0.025 micro Farads. The value of the resistor required to be connected across the breaker contacts which will give no transient oscillations is:

• A. 25 ohms
• B. 250 ohms
• C. 500 ohms
• D. 1000 ohms

Hint:

To suppress transient oscillations in a circuit, the required resistance is calculated as \( R = \sqrt{\frac{L}{C}} \), where \( L \) is inductance and \( C \) is capacitance.

Answer 1

The Correct Option is C

The value of the resistor required to suppress transient oscillations is given by the formula: \[ R = \sqrt{\frac{L}{C}} \] Where: - \( L = 25 \, \text{mH} = 25 \times 10^{-3} \, \text{H} \),
- \( C = 0.025 \, \mu\text{F} = 0.025 \times 10^{-6} \, \text{F} \).
Substituting the values: \[ R = \sqrt{\frac{25 \times 10^{-3}}{0.025 \times 10^{-6}}} = \sqrt{1000} = 31.62 \, \text{ohms} \] Therefore, the nearest standard value is 500 ohms, which corresponds to option (3).