Question

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 2.02 mm extends directly upward by 0.823 km and is coated with a 0.506 mm layer of water having resistivity 159 Ω·m. If the potential difference between the two ends of the string is 186 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

Answer:

The current is   $I = 1.1434*10^{-5}}\ A$

Explanation:

From the question we are told that

   The radius of the kite string is  $R = 2.02 mm = 0.00202 \ m$

   The  distance it extended upward is   $D = 0.823 km = 823 \ m$

   The thickness of the water layer is $d = 0.506 mm = 0.000506 \ m$

   The resistivity is  $\rho = 159\ \Omega \cdot m$

   The potential  difference is  $V = 186 MV = 186 *10^{6} \ V$

Generally the cross sectional area of the water layer is mathematically represented as

      $A = \pi r^2$

Here  r is mathematically represented as

      $r = [(R + d ) - R]$

=>   $r = [(0.00202 + 0.000506 ) - 0.00202]$

=>  $r = 0.000506$

=>     $A = 3.142 * [0.000506]^2$  

=>     $A = 8.0447*10^{-7}\ m^2$  

Generally the resistance of the water is mathematically represented as

    $R = \frac{\rho * D }{A}$

=>   $R = \frac{159 *823 }{8.0447*10^{-7}}$

=>   $R = 1.62662 * 10^{11} \ \Omega$

Generally the current is mathematically represented as

      $I = \frac{V}{R}$

=>    $I = \frac{186 *10^{6} }{1.62662 * 10^{11}}$

=>    $I = 1.1434*10^{-5}}\ A$