The inductance of single phase two wire power transmission line per kilometer gets doubled when

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Examoogle · Accepted Answer

Option C: Distance between the wires is increased as square of original distance

Explanation & Concepts:
The inductance $L$ of a single-phase two-wire transmission line depends logarithmically on the ratio of the distance between conductors ($D$) to the radius of the conductor ($r$). Because the dependence is $L \propto \ln(D/r)$, doubling the inductance requires the ratio $D/r$ to be squared, meaning $D$ must be increased as the square of the original distance.

Working Principle: The inductance of a two-wire line is given by $L = 4 	imes 10^{-7} \ln(D/r') 	ext{ H/m}$, where $r' = 0.7788r$ is the GMR of the conductor. To double the inductance $L_1$ to $L_2 = 2L_1$, we set $\ln(D_2/r') = 2 \ln(D_1/r')$. By logarithmic properties, this simplifies to $\ln(D_2/r') = \ln((D_1/r')^2)$, which implies $D_2 = (D_1^2 / r')$. Thus, the distance must increase to the square of the original value scaled by the GMR.

Key Formulas: $L = 4 	imes 10^{-7} \ln\left(\frac{D}{r'}ight) 	ext{ H/m}$ — Inductance of a single-phase two-wire line; $r' = 0.7788r$ — Geometric Mean Radius (GMR) of a solid cylindrical conductor

Key Points: Inductance is dominated by the magnetic flux linkage between the conductors.; The term $\ln(D/r')$ represents the flux linkage due to the current loop.; Increasing distance $D$ increases the loop area, thereby increasing inductance.; Logarithmic dependence implies that linear changes in $D$ result in non-linear changes in inductance.

Advantages: High spacing reduces capacitance-to-ground coupling.; Mathematical model allows precise prediction of line impedance.

Disadvantages: Increased spacing leads to higher line reactance.; Larger spacing increases the physical size and cost of the transmission towers.

Applications: Design of overhead transmission lines.; Calculation of voltage regulation and power flow in power systems.

Additional Info: Standard values: $r$ is the radius of the conductor, $D$ is the distance between conductor centers.. Option A/B are incorrect because the relationship is logarithmic, not linear or power-based in a simple ratio; specifically $L \propto \ln(D)$, not $L \propto D$.

Answer: C is correct — because inductance follows a logarithmic relationship with distance, the distance must be squared relative to the GMR factor to double the total inductance value.

Exam Tip: Always remember that in transmission line parameter calculations, internal inductance ($1/2 	imes 10^{-7}$) is constant regardless of spacing, while external inductance depends on the log of spacing.

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Option A: Distance between the wires is doubled

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Option B: Distance between the wires is increased four fold

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Option D: Radius of the wire is doubled

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