The advantages of high transmission voltage are ______

Question

Examoogle · Accepted Answer

All of these

Explanation: High transmission voltage is preferred because it significantly reduces power losses and transmission costs. By increasing the voltage, the current required to transmit a specific amount of power decreases, leading to improved overall efficiency and reduced conductor material requirements.

Working Principle: According to the power formula $P = VI \cos \phi$, for a constant power $P$, if voltage $V$ is increased, current $I$ must decrease proportionally. Since power loss ($I^2R$) is directly proportional to the square of the current, reducing $I$ leads to a drastic reduction in transmission line losses. Additionally, lower current allows for a smaller conductor cross-section (governed by Kelvin's Law), which minimizes the initial capital investment in raw materials.

Key Formulas: $P = \sqrt{3} V_L I_L \cos \phi$ — Relationship between power, voltage, and current; $P_{loss} = 3 I^2 R$ — Transmission line power loss formula; $Volume_{conductor} \propto \frac{1}{V^2}$ — Relationship between conductor volume and transmission voltage

Key Points: High voltage leads to lower line current for a given power rating.; Reduction in $I^2R$ losses enhances transmission efficiency.; Smaller conductor cross-section reduces structural weight and tower cost.; Higher voltage improves voltage regulation of the transmission line.

Advantages: Reduced line copper/aluminum losses; Increased transmission efficiency; Lower conductor material cost (smaller area); Increased power transmission capability

Disadvantages: Increased cost of insulation (transformers and switchgear); Higher risk of corona discharge and radio interference; Requirement for taller towers to maintain clearance

Applications: Long-distance bulk power transmission; Interconnection of regional power grids

Additional Info: Economic conductor size is determined by Kelvin's Law, balancing interest on capital cost vs cost of wasted energy.. Option D is valid because as $I = P / (\sqrt{3} V \cos \phi)$, decreasing $I$ allows for a smaller current-carrying capacity conductor.

Answer: C is correct — Increasing transmission voltage leads to reduced current, which minimizes $I^2R$ losses and allows for smaller conductor cross-sections.

Exam Tip: In exams, remember that while higher voltage reduces conductor cost, it drastically increases the cost of insulation and substations. This is why there is always an optimal economic voltage for every transmission length.

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