Question

A 250 kV power transmission line carrying 5.00 ✕ 102 A is hung from grounded metal...

A 250 kV power transmission line carrying 5.00 ✕ 102 A is hung from grounded metal towers by ceramic insulators, each having a 2.00 ✕ 109 Ω resistance. (a) What is the resistance to ground in ohms of 145 of these insulators? (Assume the insulators are connected in parallel.) Ω (b) Calculate the power in watts dissipated by 145 of them. W (c) What fraction of the power carried by the line is this? P Ptot =

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Answer #1

a)
Resistance to the ground when 145 insulators are connected in parallel is,
Reff = R/145
Where Reff is the combined resistance of 145 insulators with resistance R.
Reff = (2 x 109) / 145
= 1.38 x 107 Ohm

b)
Power dissipated, P = V2/Reff
= (250 x 103)2 / (1.38 x 107)
4531.25 W

c)
Total power carried by the lines, Ptot = VI
= 250 x 103 x 5 x 102
= 1250 x 105 W

Fraction of total power carried by insulators = P/Ptot
= 4531.25 W / (1250 x 105 W)
= 3.63 x 10-5

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