A 260-kv power transmission line carrying 420 A is hung from grounded metal towe

Question

A 260-kv power transmission line carrying 420 A is hung from grounded metal towers by ceramic insulators, each having a 1.00 times 10^9 -Ohm resistance (see figure below). What is the resistance to ground of 115 of these insulators? Figure caption: High-voltage (260-kv) transmission line carrying 420 A is hung from a rounded metal transmission tower. The row of ceramic insulators provide 1.00 times 10^9 Ohm of g resistance each. Calculate the power dissipated by 115 of them. what fraction of the power carried by the line is this? Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors.

Explanation / Answer

a)
Resistance of the ceramic, R = 1 x 109 Ohm.
We can consider 115 ceramics connected in parallel combination.
The net resistance, Rn = R/115 = 8.696 x 106 Ohm
Voltage between ground and the power lines = 260 x 103 V

Power dissipated, P1 = V2/Rn
= (260 x 103)2/(8.696 x 106)
= 894.01 W.

b)
Power dissipated by transmission line,
P2 = V x I
= (260 x 103) x 420
= 109.2 x 106 W

Fraction of power dissipated, P1/P2 = 894.01 W / (109.2 x 106 W)
= 8.187 x 10-6.

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