Examoogle
ExamsTest SeriesCBATRank CheckPrevious Year PapersPassBook StoreMy BooksAI Tutor
🛒0
अA
Examoogle

India's most trusted platform for competitive exam PDF books. Expert-authored, watermark-protected, instant access.

Exams & Practice
All Exams & SyllabusMock Test SeriesPrevious Year PapersPractice Questions (MCQs)Recruitment Notifications
Quick Links
Examoogle AI TutorExam NewsBook StoreMy BooksLogin / Sign Up
Support
About UsRefund PolicyPrivacy PolicyTerms of UseContact Us
© 2026 Examoogle. India's #1 competitive exam AI tutor.
🔒 SSL Secured📱 UPI Accepted🧾 GST Invoice
Examoogle

Join 60,000+ competitive exam aspirants

or with email
By continuing, you agree to ourTerms of Service&Privacy Policy
Your Cart
Subtotal₹0
Total₹0
Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Back to Practice Questions
ElectricalPower System
PrevNext

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

A

3×\sqrt{3} \times3​× post fault line voltage in kV ×\times× SC current in kA

B

3×3 \times3× pre fault line voltage in kV ×\times× SC current in kA

C

3×\sqrt{3} \times3​× pre fault line voltage in kV ×\times× SC current in kA

D

13×\frac{1}{\sqrt{3}} \times3​1​× pre fault line voltage in kV ×\times× SC current in kA

Correct Answer

⚙️ TE • Technical Concept & PrincipleElectricalPower System
Option C

3×\sqrt{3} \times3​× pre fault line voltage in kV ×\times× SC current in kA

Quick Summary:

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 3\sqrt{3}3​, the pre-fault line-to-line voltage, and the short-circuit current.

⚙️TETechnical SolutionConcept & Principle
💡 Explanation

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 3\sqrt{3}3​, the pre-fault line-to-line voltage, and the short-circuit current.

🔢 Key Formulas

MVAsc=3×VL×IscMVA_{sc} = \sqrt{3} \times V_L \times I_{sc}MVAsc​=3​×VL​×Isc​ — Calculation of short circuit MVA

Isc=VphZeqI_{sc} = \frac{V_{ph}}{Z_{eq}}Isc​=Zeq​Vph​​ — Calculation of 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 SMVA=3×Vline(kV)×Isc(kA)S_{MVA} = \sqrt{3} \times V_{line(kV)} \times I_{sc(kA)}SMVA​=3​×Vline(kV)​×Isc(kA)​.

📌 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 / Limitations
  • ▸

    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 / Uses
  • ▸

    Circuit breaker selection and sizing in power distribution networks.

  • ▸

    Protection coordination studies in substation design.

📄 Additional Information
  • ▸

    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 3\sqrt{3}3​, and Option A uses post-fault voltage, which is nearly zero at the fault point.

📊 Diagram / Illustration
Short Circuit MVA Formula√3 × V_line ×I_sc10°3 (if using Volts and Amperes)
✅

C is correct — The short-circuit MVA capacity is calculated as 3×Vpre−fault(line)×Isc\sqrt{3} \times V_{pre-fault(line)} \times I_{sc}3​×Vpre−fault(line)​×Isc​.

Core Concepts Used
Click any tag to open in AI Tutor
Symmetrical Fault Analysis Circuit Breaker Rating Power System Stability
💡 EXAM TIP

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

Related Questions

ElectricalPower System
The specified quantities of Generation bus is
ElectricalPower System
When an alternator connected to the bus-bar is shut down the bus-bar voltage will
ElectricalPower System
The current drawn by the line due to corona losses is _______
ElectricalPower System
The specified quantities of load bus are
ElectricalPower System
The advantages of high transmission voltage are ______

Discussion (0)

Loading discussion...
PrevNext