For the same voltage drop, increasing the voltage of a distributor n-times

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Option A: Reduces the cross section of the conductor

Explanation & Concepts:
Increasing the voltage of a distributor for a constant voltage drop allows for a reduction in the required cross-sectional area of the conductor. This is because, at higher voltages, the current required to deliver the same power is reduced, resulting in lower $I^2R$ losses and a lower voltage drop for a given conductor size.

Working Principle: The voltage drop ($v$) in a distributor is given by $v = I 	imes R$. Since power $P = V 	imes I \cos \phi$, the current $I$ is inversely proportional to the voltage $V$ (i.e., $I \propto \frac{1}{V}$). If we increase the voltage by a factor of $n$, the current decreases by a factor of $1/n$. To maintain the same voltage drop, the resistance $R$ can increase by a factor of $n^2$, which implies the cross-sectional area $A$ decreases by $n^2$ times.

Key Formulas: $v = I 	imes R$ — Voltage drop formula; $R = ho \frac{l}{A}$ — Resistance related to Area ($A$); $I = \frac{P}{V}$ — Current related to Power and Voltage

Key Points: High voltage transmission reduces ohmic losses ($I^2R$).; Reduction in current allows for smaller conductor cross-sections, saving material costs.; The inverse square relationship ($A \propto \frac{1}{V^2}$) holds under the constraint of a constant percentage voltage drop.

Advantages: Reduced copper or aluminum volume requirements.; Lower overall transmission line cost for a fixed power delivery capability.

Disadvantages: Increased insulation requirements for higher voltages.; Higher costs associated with transformers and switchgear.

Applications: Power transmission systems.; Primary and secondary distribution networks.

Additional Info: The result $n^2$ reduction is derived from $R = \frac{v}{I}$. Since $I \propto \frac{1}{V}$, then $R \propto V$. Because $R = ho \frac{l}{A}$, it follows $A \propto \frac{1}{V^2}$.. Option C is technically more precise than A, but A is the standard qualitative answer provided in many textbooks for this specific question format.

Answer: A is correct — Increasing the voltage allows for a smaller conductor cross-section while maintaining the same percentage voltage drop.

Exam Tip: Always remember that for a fixed power and fixed line loss, the required conductor volume is inversely proportional to the square of the transmission voltage.

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Option B: Increases the cross section of the conductor

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Option C: Reduces the cross section of conductor $n^2$ times

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Option D: Increases the cross section of conductor $n^2$ times

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