Which formula correctly expresses the symmetrical three-phase fault current in terms of line-to-line voltage and equivalent impedance?

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Prepare for the LADWP Electric Station Operator Test focusing on Circuit Breakers, Disconnects, and Transformers. Study with tailored questions and detailed explanations to enhance your knowledge and boost confidence. Ace your exam!

Multiple Choice

Which formula correctly expresses the symmetrical three-phase fault current in terms of line-to-line voltage and equivalent impedance?

Explanation:
For a symmetrical three-phase fault, the positive-sequence network governs the fault current because all three phases short together in a balanced way. The fault is driven by the line-to-line voltage, since that is the voltage directly across the shorted network when the three lines are tied together. The current that flows into the fault is then determined by Ohm’s law using the positive-sequence impedance seen from the fault location, Z_eq. The resulting current magnitude is I_sc(3φ) = √3 × V_LL / Z_eq. The factor √3 appears because line-to-line voltage relates to phase voltage by V_LL = √3 × V_phase in a balanced system, and the symmetric short-circuit path effectively uses the line-to-line quantity to drive the single equivalent impedance. As long as the same base is used for both voltage and impedance (consistency in per-unit scaling), this expression holds. Using just V_LL / Z_eq would imply the fault is driven by a single phase voltage, which is not the case for a three-phase fault. The √3 factor, arising from the relationship between line-to-line and phase quantities in a balanced network, correctly captures the driving voltage in the symmetrical fault scenario.

For a symmetrical three-phase fault, the positive-sequence network governs the fault current because all three phases short together in a balanced way. The fault is driven by the line-to-line voltage, since that is the voltage directly across the shorted network when the three lines are tied together. The current that flows into the fault is then determined by Ohm’s law using the positive-sequence impedance seen from the fault location, Z_eq.

The resulting current magnitude is I_sc(3φ) = √3 × V_LL / Z_eq. The factor √3 appears because line-to-line voltage relates to phase voltage by V_LL = √3 × V_phase in a balanced system, and the symmetric short-circuit path effectively uses the line-to-line quantity to drive the single equivalent impedance. As long as the same base is used for both voltage and impedance (consistency in per-unit scaling), this expression holds.

Using just V_LL / Z_eq would imply the fault is driven by a single phase voltage, which is not the case for a three-phase fault. The √3 factor, arising from the relationship between line-to-line and phase quantities in a balanced network, correctly captures the driving voltage in the symmetrical fault scenario.

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