How Do Changes in Dimensions Affect Electrical Resistance in Conductive Paint?

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The discussion focuses on the impact of dimensional changes on electrical resistance in conductive paint and power loss in electrical conductors. When a volume of conducting paint is poured into a cylinder with half the base area, the resistance increases to 36.8Ω due to the combined effects of reduced cross-section and increased height. Additionally, for the National Grid's 400kV supergrid conductors, the power loss per kilometer is calculated to be 7.65MW/km, assuming 100% efficiency in transformers. The calculations provided by participants highlight the importance of considering both height and cross-sectional area in resistance calculations.

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Can someone give me explanations for the answers to these questions as the book just gives small simple answers?

1)A volume of 1.2x10^-5 m^3 of conducting paint is poured into a glass cylinder of base area 3.0x10^-4 m^2. It fills the cylinder to a height of 40mm. The resistance between the top and bottom surfaces of the paint is 9.2Ω. Resistivity of the paint is 0.069 Ωm
Suppose the paint had been poured into a cylinder with half the base area. What would the value of the resistance between the top and bottom of the surfaces have been?
Books answer:36.8≈37
My answer:18.42)The power stations providing electricity for the National Grid generate electricity at 25kV. The typical demand on an average winter's day in the north-west of England is 6000MW. The current output form the generators is 240kV. Transformers raise the transmission voltage to 400kV. Assuming 100% efficiency in the transformers, the current is 15kA. Each pair of 400kV supergrid conductors has a resistance of 0.034Ωk/m.
How much power is lost per kilometre heating the conductors?
Books answer:7.65MW k/m
My answer:5.988 MW k/m
 
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It would be easier to find your errors if you included the calculation here.

1) did you take the doubled height into account? Both this and the reduced cross-section increase the resistance by a factor of 2, for a total of 4*9.2Ω=36.8Ω.
2) this has a strange unit for the resistance. Ωk/m? Ω/km?
Assuming the second, did you calculate the voltage drop in 1km? Multiply it with the current and you get the same result as the book.
 

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