Examoogle
ExamsTest SeriesCBATRank CheckPrevious Year PapersPassBook StoreMy BooksAI Tutor
ЁЯЫТ0
рдЕA
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
┬й 2026 Examoogle. India's #1 competitive exam AI tutor.
ЁЯФТ SSL SecuredЁЯУ▒ UPI AcceptedЁЯз╛ GST Invoice
Examoogle

Join 60,000+ competitive exam aspirants

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

Inductance of single phase two wire line is given by

A

$0.4 \ln(D/r) \text{ mH/km}$

B

$0.55 \ln(D/r) \text{ mH/km}$

C

$0.4 \ln(r/D) \text{ mH/km}$

D

$0.55 \ln(r/D) \text{ mH/km}$

Correct Answer

тЪЩя╕П TE тАв Technical Concept & PrincipleElectricalPower System
Option A

$0.4 \ln(D/r) \text{ mH/km}$

Quick Summary:

The inductance of a single-phase two-wire transmission line is derived from the flux linkages of the conductors ┬╖ For two parallel conductors of radius 'r' separated by a distance 'D', the total inductance per unit length is the sum of internal and external inductances, resulting in a value of approximately $0.4 \ln(D/r)$ mH/km.

тЪЩя╕ПTETechnical SolutionConcept & Principle
ЁЯТб Explanation

The inductance of a single-phase two-wire transmission line is derived from the flux linkages of the conductors ┬╖ For two parallel conductors of radius 'r' separated by a distance 'D', the total inductance per unit length is the sum of internal and external inductances, resulting in a value of approximately $0.4 \ln(D/r)$ mH/km.

ЁЯФв Key Formulas

L=4├Ч10тИТ7lnтБб(DrтА▓)┬аH/mL = 4 \times 10^{-7} \ln(\frac{D}{r'}) \text{ H/m}L=4├Ч10тИТ7ln(rтА▓DтАЛ)┬аH/m тАФ Fundamental inductance formula using GMR (rтА▓r'rтА▓)

L=0.4lnтБб(Dr)┬аmH/kmL = 0.4 \ln(\frac{D}{r}) \text{ mH/km}L=0.4ln(rDтАЛ)┬аmH/km тАФ Practical engineering approximation

тЪЩя╕П Working Principle

The total inductance LLL is calculated by considering the flux linkage of each conductor ┬╖ The formula L=4├Ч10тИТ7lnтБб(D/rтА▓)┬аH/mL = 4 \times 10^{-7} \ln(D/r') \text{ H/m}L=4├Ч10тИТ7ln(D/rтА▓)┬аH/m is converted to mH/km, where rтА▓r'rтА▓ is the geometric mean radius (GMR) of the conductor (rтА▓=reтИТ1/4тЙИ0.7788rr' = re^{-1/4} \approx 0.7788rrтА▓=reтИТ1/4тЙИ0.7788r) ┬╖ Substituting these constants leads to the standard engineering approximation $0.4 \ln(D/r)$ mH/km.

ЁЯУМ Key Points
  • тЦ╕

    Inductance increases logarithmically as the distance between conductors (D) increases.

  • тЦ╕

    Inductance decreases as the radius (r) of the conductor increases.

  • тЦ╕

    The factor 0.7788 represents the internal flux linkage effect accounted for in the conversion from rтА▓r'rтА▓ to rrr.

  • тЦ╕

    Transmission line inductance is a key parameter for calculating voltage drops and stability.

тЬЕ Advantages
  • тЦ╕

    Provides a simple logarithmic relationship for system design.

  • тЦ╕

    Essential for modeling impedance in power systems.

тЭМ Disadvantages / Limitations
  • тЦ╕

    Assumes non-magnetic conductors and uniform current distribution.

  • тЦ╕

    Becomes complex with bundles or multiple-phase configurations.

ЁЯЫая╕П Applications / Uses
  • тЦ╕

    Design of distribution and transmission lines.

  • тЦ╕

    Calculation of power factor and voltage regulation.

ЁЯУД Additional Information
  • тЦ╕

    The constant 0.4 arises from (2├Ч2├Ч10тИТ7)├Ч1000(2 \times 2 \times 10^{-7}) \times 1000(2├Ч2├Ч10тИТ7)├Ч1000 to convert Henries per meter to mH/km.

  • тЦ╕

    Option C and D are incorrect as the argument of the natural logarithm must be greater than 1 (D > r) for a positive inductance value.

ЁЯУК Diagram / Illustration
Inductance FormulaL = 0.4 ln(D/r) mH/kmD = Spacing between centersr = Radius of conductor
тЬЕ

A is correct тАФ The inductance of a single-phase two-wire line is given by the expression $0.4 \ln(D/r)$ mH/km.

Core Concepts Used
Click any tag to open in AI Tutor
Flux Linkages Geometric Mean Radius (GMR) Inductive Reactance
ЁЯТб EXAM TIP

Remember that capacitance formula uses a similar logarithmic form but involves lnтБб(D/r)\ln(D/r)ln(D/r) in the denominator, highlighting the inverse relationship between L and C parameters in transmission lines.

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