Question

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?

Answer 1

Answer:

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

$E=34.5\ J$ energy dissipated.

Explanation:

Given:

• time for which the current lasted, $t=43\times 10^{-6}\ s$

• potential difference between the feet, $V=21000\ V$

• resistance between the feet, $R=550\ \Omega$

Now, from the Ohm's law we have:

$I=\frac{V}{R}$

$I=\frac{21000}{550}$

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

Energy dissipated in the body:

$E=I^2.R.t$

$E=38.181^2\times 550\times 43\times 10^{-6}$

$E=34.5\ J$