Question: Slide flash. Figure indicates one reason no one should stand under a tree during a lightning storm. If lightning comes down the side of the tree, a portion can jump over to the person, especially if the current on the tree reaches a dry region on the bark and thereafter must travel through air to reach the ground. In the figure, part of the lightning jumps through distance in air and then travels through the person (who has negligible resistance relative to that of air). The rest of the current travels through air alongside the tree, for a distance h. Ifand the total current is, what is the current through the person?

The ratio of distance and height is,$\frac{d}{h}=0.4$.

The total current is,$I=5000 \text{A}$.

Here, we have to use equation of Kirchhoff’s current law and the equation of the resistance related with the length, the resistivity and the area.

Formula:

$$\begin{gathered} \mathit{V}\mathbf{}\mathbf{=}\mathbf{}\mathit{I}\mathit{R} \\ \mathit{R}\mathbf{=}\frac{\mathbf{\rho }\mathbf{L}}{\mathbf{A}} \end{gathered}$$

Here, let’s assume V₁ is the voltage through the left side and through the right side.

Both are parallel. So, the voltage is same.

$V_1=V_2$

Means,

$I_1{R}_1=I_2{R}_2$ …(1)

But we know,

$R=\frac{\rho L}{A}$

So,

$R_1=\frac{\rho d}{A}$

And

$R_2=\frac{\rho h}{A}$

Substitute all the value in the equation (1),

$$\begin{gathered} I_{1}d=I_{2}h \\ {I}_2=I_1\times \frac{d}{h} \end{gathered}$$

Substitute all the value in the above equation.

$I_2=0.4\times I_2$

But according to Kirchhoff’s current law,

$I_1+I_2=5000$

Now,

$$\begin{gathered} I_1+0.4\times I_1=5000 \text{A} \\ 1.4\times I_1=5000 \text{A} \\ {I}_1=3571.43 \text{A} \\ {I}_1\simeq 3600 \text{A} \end{gathered}$$

Hence the current through the person is, 3600A.