How many MVars are gained when energizing a capacitor rated at 100 MVar on a 345 kV line at 90% of normal voltage?

How many MVars are gained when energizing a capacitor rated at 100 MVar on a 345 kV line at 90% of normal voltage?

• A. 50 MVars
• B. 81 MVars
• C. 90 MVars
• D. 30 MVars

Answer: (B)

When energizing a capacitor, the reactive power (measured in MVars) that it can provide to the system is influenced by the voltage level at which it is operated. A capacitor's rated reactive power is typically based on its maximum potential at nominal voltage conditions. In this case, the capacitor is rated at 100 MVars, and the system's voltage is at 90% of the normal voltage. To find how many MVars are gained at 90% of the voltage, the formula used is: \[ \text{MVars} = \text{Rated MVars} \times \left(\frac{V}{V_{\text{nom}}}\right)^2 \] Substituting the values into the formula, we get: \[ \text{MVars} = 100 \times \left(0.90\right)^2 = 100 \times 0.81 = 81 \text{ MVars} \] This means that when the capacitor is energized at 90% of the nominal voltage, it will supply approximately 81 MVars of reactive power to the system. This adjustment is crucial, as it reflects the reality that the capacitance effect diminishes when the operating voltage is

When energizing a capacitor, the reactive power (measured in MVars) that it can provide to the system is influenced by the voltage level at which it is operated. A capacitor's rated reactive power is typically based on its maximum potential at nominal voltage conditions. In this case, the capacitor is rated at 100 MVars, and the system's voltage is at 90% of the normal voltage.

To find how many MVars are gained at 90% of the voltage, the formula used is:

[

\text{MVars} = \text{Rated MVars} \times \left(\frac{V}{V_{\text{nom}}}\right)^2

]

Substituting the values into the formula, we get:

[

\text{MVars} = 100 \times \left(0.90\right)^2 = 100 \times 0.81 = 81 \text{ MVars}

]

This means that when the capacitor is energized at 90% of the nominal voltage, it will supply approximately 81 MVars of reactive power to the system. This adjustment is crucial, as it reflects the reality that the capacitance effect diminishes when the operating voltage is