Question

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 0.823 km and is coated with a 0.517 mm layer of water having resistivity 168 Ω·m. If the potential difference between the two ends of the string is 165 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).

Answer 1

current through water layer is given as 0.0090051 A

Step-by-step explanation:

Given data:

Radius of kite string is 2.10 mm

Thickness of water coating 0.517 mm

distance of kite in upward direction = 0.823 km

Resistivity = 168 ohm. m

Potential difference is 165 MV

Cross-sectional area of water

`A = pi*((2.61*10^(-3)m)^2 - (2.10*10^(-3))^2`

`= 7.546*10^(-6) m^2`

Resistance of water
`R = resistivity * (length\ L )/(area\ A)`

`= 168* (823)/(7.5465*10^(-6))`

`= 1.832 * 10^(10) Ohm`

Current through water
`I =( V )/( R)`

`= (165* 10^6)/(1.832* 10^(10)) Ohm`

= 0.0090051 A