Electrical Power Transmission MCQs Part 22

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Electrical Power Transmission MCQs Part 22

The favorable economic factor involved in power transmission at high voltage is

The favorable economic factor involved in power transmission at high voltage is:

  1. Smaller size of generation plant
  2. Decreased insulation required for line conductors
  3. Reduction of conduction cross-section area
  4. Increased I²R losses

Correct answer: 3. Reduction of conduction cross-section area

EHV cables are filled with thin oil under pressure in order to

EHV cables are filled with thin oil under pressure in order to:

  1. Prevent entry of moisture
  2. Prevent formation of voids
  3. To provide insulation
  4. To strengthen cable conductor

Correct answer: 2. Prevent formation of voids

Range of accelerating factor is

Range of accelerating factor is

  1. 1 to 1.5
  2. 1.6 to 1.8
  3. 10.8 to 11.98
  4. 50 to 100

Correct answer: 2. 1.6 to 1.8

Explanation: Accelerating factor is used for reducing number of iterations using Gauss-Siedel method. The range of accelerating factor is between 1.6 to 1.8.

The total diameter of a 3 layer ACSR conductor whose each strand has diameter d, then find the total diameter of ACSR conductor

The total diameter of a 3 layer ACSR conductor whose each strand has diameter d, then find the total diameter of ACSR conductor

  1. d
  2. 2d
  3. 3d
  4. 5d

Correct answer: 4. 5d

Explanation:

Total diameter of ACSR conductor D = (2X – 1) X d

Where,

X = layer number

d = diameter of each strand

Therefore,

Total diameter of ACSR conductor D = (2 X 3 – 1) X d = 5d

The internal inductance of an ACSR conductor is 0.05 mH/Km for µr = 1. If the ACSR is replaced with another having µr = 2, then find the internal inductance

The internal inductance of an ACSR conductor is 0.05 mH/Km for µr = 1. If the ACSR is replaced with another having µr = 2, then find the internal inductance:

  1. 0.1 mH/Km
  2. 0.3 mH/Km
  3. 0.5 mH/Km
  4. 0.7 mH/Km

Correct answer: 1. 0.1 mH/Km

Explanation:

Formula for internal inductance Lint = µr/(2 x 10-7)

Therefore:

Lint ∝ µr

µr = Relative permeability

Lint2 = Lint1 X (µr2 / µr1) = 0.05 * 2/1 = 0.1 mH/Km

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