Asked by AEC
A thunderstorm with a potential difference of ten million Volts between two clouds produces a lightning stroke of ten thousand Amperes at its peak with a rise time of tenmicroseconds and a fall time of one hundred microseconds. How many Coulombs does it carry? What are the peak power and the total energy of the lightning stroke?
Answers
Answered by
Damon
we need to find the total charge passed during the rise and fall
during rise
i = {10^4amps /[10^-5 s]}t
area under that line (the triangle) is in coulombs
Q = .5*10^4*10^-5 = .05 Coulombs
during fall
Q = .5*10^4 *10^-4 = .5 Coulombs
peak power = volts * peak amps
10^7 * 10^4 = 10^11 watts (yikes)
total energy = integral of Vi dt assuming V is constant that is V int i dt
but we know integral i dt = .05 + .5 =
.55
so 10^7 Volts * .55 coulombs total energy
during rise
i = {10^4amps /[10^-5 s]}t
area under that line (the triangle) is in coulombs
Q = .5*10^4*10^-5 = .05 Coulombs
during fall
Q = .5*10^4 *10^-4 = .5 Coulombs
peak power = volts * peak amps
10^7 * 10^4 = 10^11 watts (yikes)
total energy = integral of Vi dt assuming V is constant that is V int i dt
but we know integral i dt = .05 + .5 =
.55
so 10^7 Volts * .55 coulombs total energy
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