Three phase short circuit MVA to be interrupted by a circuit breaker in a power system is given by

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Option C: $\sqrt{3} 	imes$ pre fault line voltage in kV $	imes$ SC current in kA

Explanation & Concepts:
The interrupting capacity of a circuit breaker in a three-phase system is defined by its ability to break the short-circuit MVA at a given voltage level. This is calculated using the product of $\sqrt{3}$, the pre-fault line-to-line voltage, and the short-circuit current.

Working Principle: During a symmetrical three-phase fault, the power system delivers maximum current to the fault point. The total apparent power (MVA) is calculated based on the system voltage prior to the fault and the steady-state fault current. Since the breaker must interrupt the total fault energy, the rating is defined as $S_{MVA} = \sqrt{3} 	imes V_{line(kV)} 	imes I_{sc(kA)}$.

Key Formulas: $MVA_{sc} = \sqrt{3} 	imes V_L 	imes I_{sc}$ — Calculation of short circuit MVA; $I_{sc} = \frac{V_{ph}}{Z_{eq}}$ — Calculation of short circuit current

Key Points: The voltage used is the rated pre-fault line-to-line voltage.; Short circuit MVA is a standard metric used to determine the interrupting rating of circuit breakers.; This formula assumes a balanced three-phase system under symmetrical fault conditions.

Advantages: Provides a simple, universal metric for comparing circuit breaker capabilities.; Directly relates fault energy to system voltage levels.

Disadvantages: Does not account for the DC offset component present in the transient period of a fault.; Calculations are based on steady-state values and may underestimate initial mechanical stress.

Applications: Circuit breaker selection and sizing in power distribution networks.; Protection coordination studies in substation design.

Additional Info: The term 'pre-fault' voltage is used because it represents the system condition before the voltage collapses due to the short circuit.. Option B is incorrect as it uses a factor of 3 instead of $\sqrt{3}$, and Option A uses post-fault voltage, which is nearly zero at the fault point.

Answer: C is correct — The short-circuit MVA capacity is calculated as $\sqrt{3} 	imes V_{pre-fault(line)} 	imes I_{sc}$.

Exam Tip: Remember that for symmetrical faults, always use line-to-line voltage with $\sqrt{3}$, but if using phase voltage, the factor becomes 3.

Examoogle · Answer

Option A: $\sqrt{3} 	imes$ post fault line voltage in kV $	imes$ SC current in kA

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Option B: $3 	imes$ pre fault line voltage in kV $	imes$ SC current in kA

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Option D: $\frac{1}{\sqrt{3}} 	imes$ pre fault line voltage in kV $	imes$ SC current in kA

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