Kiting during a storm. The legend that Benjamin Franklin flew a kite as a storm approached is only a legend — he was neither stupid nor suicidal. Suppose a kite string of radius 2.10 mm extends directly upward by 0.802 km and is coated with a 0.520 mm layer of water having resistivity 155 Ω·m. If the potential difference between the two ends of the string is 176 MV, what is the current through the water layer? The danger is not this current but the chance that the string draws a lightning strike, which can have a current as large as 500 000 A (way beyond just being lethal).

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Asked Oct 14, 2019
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Kiting during a storm. The legend that Benjamin Franklin flew a kite as a storm approached is only a legend — he was neither stupid nor suicidal. Suppose a kite string of radius 2.10 mm extends directly upward by 0.802 km and is coated with a 0.520 mm layer of water having resistivity 155 Ω·m. If the potential difference between the two ends of the string is 176 MV, what is the current through the water layer? The danger is not this current but the chance that the string draws a lightning strike, which can have a current as large as 500 000 A (way beyond just being lethal).

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Expert Answer

Step 1

Consider the radius of the string be r, the length of the string be l, the resistivity of the water be ρ, the thickness of the water layer be d, the current through the water layer be I, the resistance of the wire be R, and the potential difference be V.

 

The given values are,

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d 0.520 mm = 0.520 x 103 m r 2.10 mm 2.10 x 103 m 1 0.802 km = 0.802 x 103 m =802 m V 176 MV 176 x 106 V p= 155 .m

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Step 2

The radius of the string along with water layer can be calculated as,

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r'=rd =(2.10x103 m)+(0.520 x 103 m) 2.620 x103 m

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Step 3

The area of the cross section of the...

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A=(r ((2.620x10 m)-(2.10x10 m)) m = 7.710 x10

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