If system reactive power is 1 MVAR and short circuit capacity is 81.34 MVA/phase then the ratio of r

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If system reactive power is 1 MVAR and short circuit capacity is 81.34 MVA/phase then the ratio of receiving end to sending end voltage is

Examoogle · Accepted Answer

0.99

Explanation: Given: System reactive power Q = 1 MVAR, Short circuit capacity S_sc = 81.34 MVA/phase.

Formula Used: $$Q = \frac{V_s^2 - V_s V_r}{X} = \frac{V_s(V_s - V_r)}{X}, \quad S_{sc} = \frac{V_s^2}{X} \implies \frac{V_r}{V_s} = 1 - \frac{Q}{S_{sc}}$$

Steps: Identify Relationship: The change in voltage due to reactive power flow in a system is related to the short circuit capacity ($S_{sc}$). The voltage drop can be approximated using the ratio of reactive power to short circuit capacity.. Rearrange for Ratio: We need the ratio of receiving end voltage ($V_r$) to sending end voltage ($V_s$). Rearranging the approximation formula gives:. Substitute Values: Substitute $Q = 1$ MVAR and $S_{sc} = 81.34$ MVA/phase into the equation.. Calculate Final Result: Performing the division $1 / 81.34 \approx 0.0123$, then subtracting from 1.

Answer: A is correct because the ratio of receiving end to sending end voltage, calculated as $1 - (Q / S_{sc})$, yields approximately 0.99.

Exam Tip: This concept is a cornerstone of Power System Analysis, specifically in understanding grid strength; a higher short circuit capacity implies a 'stiffer' grid with less voltage fluctuation.

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