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

In summary, the first question deals with conducting paint being poured into a glass cylinder. The resistance between the top and bottom surfaces of the paint is 9.2Ω and the resistivity is 0.069 Ωm. If the paint had been poured into a cylinder with half the base area, the resistance between the two surfaces would have been 18.4Ω. The book's answer is 36.8Ω.The second question relates to power stations generating electricity at 25kV and a typical demand of 6000MW. The transformers increase the transmission voltage to 400kV with a current of 15kA. The resistance of each pair of 400kV supergrid conductors
  • #1
Delpo
4
0
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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  • #2
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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