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A person in bare feet is standing under a tree during a thunderstorm, seeking shelter from the rain. A lightning strike hits the tree. A burst of current lasting 43 µs passes through the ground; during this time the potential difference between his feet is 21 kV. If the resistance between one foot and the other is 550 Ω,

what is the current through his body and how much energy is dissipated in his body by the lightning?

1 Answer

1 vote

Answer:


I=38.181\ A is the current through the body of the man.


E=34.5\ J energy dissipated.

Step-by-step explanation:

Given:

  • time for which the current lasted,
    t=43* 10^(-6)\ s
  • potential difference between the feet,
    V=21000\ V
  • resistance between the feet,
    R=550\ \Omega

Now, from the Ohm's law we have:


I=(V)/(R)


I=(21000)/(550)


I=38.181\ A is the current through the body of the man.

Energy dissipated in the body:


E=I^2.R.t


E=38.181^2* 550* 43* 10^(-6)


E=34.5\ J

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