Two arrangements of conductors are proposed for a 3-phase transmission line: one with equilateral spacing of 4

Question

Two arrangements of conductors are proposed for a 3-phase transmission line: one with equilateral spacing of 4 m and the other a flat with 4 m between the conductors · The conductor diameter in each case is 2 cm · Assuming that the line is transposed in both cases · Which one of the following statements would be true?

(Cₙ = capacitance in F/m line to neutral, L = inductance in H/m per phase)

Examoogle · Accepted Answer

Option B: C_n1 > C_n2 and L₁ < L₂ Explanation & Concepts: For a 3-phase line, the Geometric Mean Distance (GMD) for equilateral spacing is $D_{eq} = d = 4 ext{ m}$, whereas for flat spacing, $D_{eq} = \sqrt[3]{d imes d imes 2d} = \sqrt[3]{2} imes d \approx 1.26d = 5.04 ext{ m}$. Since inductance $L$ is proportional to $\ln(D_{eq}/GMR)$ and capacitance $C_n$ is inversely proportional to $\ln(D_{eq}/r)$, a larger $D_{eq}$ results in higher inductance and lower capacitance. Working Principle: The inductance of a transposed line depends on the GMD between phases. Equilateral spacing results in a smaller $D_{eq}$ compared to flat spacing, leading to lower inductance. Conversely, capacitance is inversely related to the logarithmic term involving GMD; thus, the lower $D_{eq}$ in equilateral spacing leads to a higher capacitance compared to the flat spacing arrangement. Key Formulas: $L = 2 imes 10^{-7} \ln(\frac{D_{eq}}{r'}) ext{ H/m}$ — Inductance per phase; $C_n = \frac{2\pi \epsilon_0}{\ln(D_{eq}/r)} ext{ F/m}$ — Capacitance to neutral Key Points: Equilateral spacing minimizes $D_{eq}$ for a fixed minimum distance $d$.; Inductance increases with $D_{eq}$ due to the logarithmic term.; Capacitance decreases as $D_{eq}$ increases due to the inverse logarithmic relationship.; Transposition ensures balanced parameters in asymmetric (flat) configurations. Advantages: Equilateral spacing provides lower reactive impedance ($X_L$).; Equilateral spacing provides higher shunt capacitance ($C_n$). Disadvantages: Flat spacing is physically easier to construct on standard towers.; Increased $D_{eq}$ in flat spacing raises line voltage drop due to increased reactance. Applications: High voltage transmission lines; Power system grid design Additional Info: For Equilateral: $D_{eq} = 4 ext{ m}$. For Flat: $D_{eq} = (4 imes 4 imes 8)^{1/3} = 5.04 ext{ m}$.. Option B is correct because $L_1 C_{n2}$ follow directly from $D_{eq1} Answer: B is correct — Equilateral spacing yields a smaller geometric mean distance than flat spacing, resulting in lower inductance and higher capacitance. Exam Tip: Always verify if the line is transposed; if not, you must use GMD calculations to find the equivalent single-phase representation.

Examoogle · Answer

Option A: Cn1 = Cn2 and L1 > L2

Examoogle · Answer

Option C: Cn1 n2 and L1 > L2

Examoogle · Answer

Option D: Cn1 > Cn2 and L1 = L2

Feature	Equilateral	Flat
Geometric Mean Distance ( $D_{eq}$ )	d (4 m)	1.26d (5.04 m)
Inductance (L)	Lower	Higher
Capacitance ( $C_n$ )	Higher	Lower

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