Question

In distribution systems, the size of conductor is determined by using:

• A. Faraday's law
• B. Kelvin's law
• C. Ohm's law
• D. Kirchhoff's law

Hint:

Kelvin's Law optimizes costs theoretically, but practical conductor selection must also consider physical constraints like maximum current carrying capacity (thermal limit), voltage drop limits, and mechanical corona thresholds.

Answer 1

The Correct Option is B

Concept: Conductor selection in electrical distribution and transmission networks balancing engineering costs and performance. * Kelvin's Law provides an economic criterion for determining the optimal conductor cross-sectional area by balancing annual capital costs against annual energy losses. * The total annual operating cost of a conductor can be split into two main components: 1. Annual Charge on Capital Cost ($C_1$): Covers interest and depreciation on the initial investment in the conductor material. This cost is directly proportional to the conductor size ($A$): \[ C_1 = P_1 \cdot A \] 2. Annual Cost of Wasted Energy ($C_2$): Accounts for the financial cost of $I^2R$ power losses over the year. Since resistance is inversely proportional to area ($R \propto \frac{1}{A}$), this cost varies inversely with conductor size: \[ C_2 = \frac{P_2}{A} \]

Step 1: Apply optimization calculus to find the minimum cost.
The total combined annual cost ($C_{\text{total}}$) is the sum of both components: \[ C_{\text{total}} = C_1 + C_2 = P_1 \cdot A + \frac{P_2}{A} \] To find the conductor area that minimizes the total cost, take the derivative with respect to area $A$ and set it to zero: \[ \frac{dC_{\text{total}}}{dA} = P_1 - \frac{P_2}{A^2} = 0 \] \[ P_1 = \frac{P_2}{A^2} \implies P_1 \cdot A = \frac{P_2}{A} \] \[ C_1 = C_2 \]

Step 2: Formulate Kelvin's Law.
The derivative shows that total cost is minimized when the annual interest and depreciation on the conductor material equals the annual cost of energy lost in the line. This economic principle is known as Kelvin's Law. Therefore, Option (B) is the correct choice.