Attached is a 30-seconds exposure picture of lightning across the night sky abov
ID: 1418195 • Letter: A
Question
Attached is a 30-seconds exposure picture of lightning across the night sky above Canonsburg, PA from April 2006. (Original picture found in USA Today paper, Monday June 21, 2010.) Properties of a lightning bolt, such as shown in the picture, are estimated to be as follows: cloud to ground length L = 5 miles bolt diameter d = 0.012 m potential of cloud relative to earth ground V = 3 MV resistivity highly ionized lightning bolt rho = 12 Times 10^-18 m (- 7 times greater than copper at 20 degrees C) temperature of bolt as high as 20, 000 degrees C Calculate the following electrical parameters of a lightning bolt based upon the assumed properties of a lightning bolt given above. Electrical field E within conducting lightning bolt. Resistance R of a 5 mile bolt. Current I within bolt. Current density J within bolt. Power P delivered by bolt to earth ground. Energy E delivered to earth ground in t = 30 mu s. E = V/m R = Ohm I = KA (k = 10^3) J = GA/m^2 (k = 10^9) P = GW (G = 10^9) E = MJ (M = 10^6) (Perform calculations on back page and record final answer in RESULTS above to three significant figure and in SI units requested using symbols for power of ten.)Explanation / Answer
a) electric field = 3 * 106/8046.72
= 372.82 V/m
b) resistance = (12 * 10-8 * 8046.72)/(3.14 * 0.006 * 0.006)
= 8.542 ohm
c) current = 3 * 106/8.542
= 351.2 kA
d) current density = (351.2 * 1000)/(3.14 * 0.006 * 0.006)
= 3.1068 GA/m2
e) power = (3 * 106)2/8.542
= 1053.61 GW
f) Energy = 1053.61 * 109 * 30 * 10-6
= 31.608 MJ
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