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In a transmission system, the weight of copper used is proportional to

$Voltage²$

$Voltage$

$\frac{1}{Voltage²}$

$\frac{1}{Voltage}$

Correct Answer

⚙️ TE • Technical Concept & PrincipleElectricalPower System

Option C

$\frac{1}{Voltage²}$

Quick Summary:

In an electrical power transmission system, the volume (and hence weight) of copper required for the conductors is inversely proportional to the square of the transmission voltage for a constant power delivery and power loss. Increasing the voltage allows for lower current, which drastically reduces the conductor cross-sectional area needed.

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

🔢 Key Formulas

$P = VI \cos \phi$ — Real power transmission formula

$Volume_{Cu} \propto \frac{1}{V²}$ — Relation between copper volume and transmission voltage

⚙️ Working Principle

For a fixed power transmission $P = VI \cos \phi$ , the current $I$ is given by $I = \frac{P}{V \cos \phi}$ . The power loss $L = I² R$ where $R = \rho \frac{l}{A}$ . Since $A = \frac{\rho l}{R}$ and $R = \frac{L}{I²}$ , substituting $I$ results in the conductor area $A$ being proportional to $I²$ , which corresponds to $\frac{1}{V²}$ . Thus, for a fixed percentage power loss, the copper volume $V_{cu} = A \times l$ is proportional to $V^{-2}$ .

📌 Key Points

▸
Higher transmission voltage leads to lower line currents for the same power.
▸
Lower currents allow for smaller cross-sectional area of conductors (cables).
▸
Reduction in area directly translates to a reduction in weight and cost of copper.
▸
Higher voltage transmission improves transmission efficiency and reduces $I^2R$ losses.

✅ Advantages

▸
Significant reduction in transmission line cost
▸
Reduced $I^2R$ line losses
▸
Improved voltage regulation

❌ Disadvantages / Limitations

▸
Increased insulation requirements for high voltage
▸
Higher cost of transformers and switchgear
▸
Increased corona losses at very high voltages

🛠️ Applications / Uses

▸
High Voltage Direct Current (HVDC) transmission
▸
Long-distance Extra High Voltage (EHV) AC transmission

📄 Additional Information

▸
The inverse square relationship holds true only when the percentage of power loss and the length of the line remain constant.
▸
Option A is incorrect as it describes the relationship for voltage drop capability, not conductor weight.
▸
Option B and D imply a linear relationship which fails to account for the square dependency of resistive power loss.

📊 Diagram / Illustration

✅

C is correct — The weight of copper required is inversely proportional to the square of the transmission voltage, $W \propto \frac{1}{V²}$ .

Core Concepts Used

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Ohm's Law Joule's Law of Heating Transmission Line Economics

💡 EXAM TIP

Always remember that doubling the transmission voltage reduces the required copper weight by a factor of four, which is a key economic driver in power grid design.

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