Transmission Line Parameters - Free MCQ Practice Test with solutions, GATE

Concept:
The ABCD parameters are given by:
V₁ = AV₂ - BI₂
I₁ = CV₂ - DI₂
where, V₁ and I₁ are the input voltage and current
V₂ and I₂ are the output voltage and current
Case 1: When output current (I₂) = 0
A = V₁/V₂ = Voltage ratio
C = I₁/V₂ = Admittanc
Case 2: When output voltage (V₂) = 0

Hence, option 2 is correct.

Concept:
For a triangle configuration of conductors , the mutual-GMD is given by:
GMD = (D₁₂ × D₂₃ × D₃₁)^1/3
Calculation:
Given;  D₁₂ = 2.5 m, D₂₃ = 4.5 m, D₃₁ = 2 m
GMD = (2.5 × 4.5 × 2)^1/3
GMD = 2.82 m

Series Resistance:

• The primary source of real power losses incurred in a transmission system is due to the resistance of the conductors.
• Power loss is directly proportional to the square of the RMS current traveling through the line.

ABCD Parameter:
We know that the ABCD parameter of the transmission line

A = D
AD – BC = 1
Conditions of reciprocity and symmetry in terms of different two-port parameters are:

Self GMD or GMR:

• Self GMD is also called GMR. GMR stands for Geometrical Mean Radius. 
• GMR is calculated for each phase separately.
• self-GMD of a conductor depends upon the size and shape of the conductor
• GMR is independent of the spacing between the conductors.

GMD:

• GMD stands for Geometrical Mean Distance.
• It is the equivalent distance between conductors.
• GMD depends only upon the spacing 
• GMD comes into the picture when there are two or more conductors per phase.​

The inductance of the single-phase two-wire line is

GMD = Mutual Geometric Mean Distance = D
GMR = 0.7788r
r = Radius of the conductor​
The capacitance between two conductors is

In the calculation of the capacitance, the inner radius of the conductor not considered
Therefore, The self GMD method is used to evaluate Inductance only.

From configuration:
(GMR)₁ = (GMR)₂ = (GMR)₃ = (GMR)₅ = (GMR)₆ = (GMR)₇   

(GMR)₄ =(D₄₄ × D₄₁ × D₄₂ × D₄₃ × D₄₅ × D₄₆ × D₄₇)^1/7
(GMR)₄ = (.7788r × 2r × 2r × 2r × 2r × 2r × 2r)^1/7
(GMR)₄ = 1.747r 
(GMR)₁ = (D₁₁ × D₁₂ × D₁₃ × D₁₄ × D17 × D₁₅ × D₁₆)^1/7

 (GMR)₁ =(.7788r × 2r × 2r × 2r × 4r × 2√3r × 2√3r)^1/7
(GMR)₁ = 2.257r
GMR=(1.747r × (2.257r)⁶)^1/7
GMR = 2.175r

= 8.845 cm = 0.08845 m

= 6.96 m
Inductance per phase = 2 × 10^-7  
= 8.732 × 10^-7 H/m
= 0.8732 mH/km. 

Loop inductance of a single-phase two-wire line
Considered a single-phase line consisting of two conductors (phase and neutral) a and b of equal radius r. They are situated at a distance D meters. The cross sections of conductors are shown in the diagram below.

Let the current flow in the conductors are opposite in direction so that one becomes the return path for the other.
The flux linkages of conductor ‘a’ is given by the formula:

Inductance per conductor in a three-phase line (symmetrical)
Let the spacing between the conductors be D and the radius of each conductor, r. The flux linkages of conductor a is given by the equation:

The flux linkages of the conductor ‘a’ is given by the formula:

Where D_aa = r', 
D_ab = D_bc =  D_ac = D

Now from 3 -ϕ,3 wire syste
I_a + I_b + I_c = 0
I_a = - (I_b + I_c)

Where La is the inductance per conductor in a three-phase line 
From equations (1) and (2),
Inductance per conductor in a 3-phase line = (1 / 2) times loop inductance in 1 phase.

ABCD parameters:
ABCD parameters are generalized parameters of the transmission line. Given as
V_S = A V_R + B I_R
I_S = C V_R + D I_R
In matrix form,

V_S = Sending end voltage
V_R = Receiving end voltage
I_S = Sending end current
I_R = Receiving end current
A, B, C, D = T-parameter of transmission line
Calculation:
Given:
V_S = 400 ∠0° kV 
A = D = 0.8 ∠1°
The system is under no-load condition
So, I_R = 0, ⇒ V_S = AV_R
The receiving end voltage under no-load condition

V_R = 500 ∠-1°  kV 

Concept:
The capacitance between the conductors of a 1-ϕ, 2- long solid conductors transmission line:

D = distance of separation between two long solid conductors
r = radius of each long solid conductor 
q_A, q,  = Charge across A and B conductors
The standard result for the capacitance between the conductors
C_AB = q_A / V_AB 
= q_B / V_BA 
= π ϵ_r ϵ_∘ / ln(D / r)       .....(i 
Calculations:
ϵ_∘ = 8.85 × 10^-12 
ϵ_r  = 1 (for air)
π = 3.14
ln (D / r) = 2.303 log (D / r)
From equation (i)
C_AB = (3.14 × 8.85 × 10^-12 × 1) / (2.303 log (D / r)) F/m