Question

The conducting path between the right hand and the left hand can be modeled as a 9.0-cm-diameter, 140-cm-long cylinder. The average resistivity of the interior of the human body is 5.5 Ωm . Dry skin has a much higher resistivity, but skin resistance can be made negligible by soaking the hands in salt water. If skin resistance is neglected, what potential difference between the hands is needed for a lethal shock of 100 \rm mA across the chest? Your result shows that even small potential differences can produce dangerous currents when the skin is wet.
Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

The potential difference is 121.069 V

Solution:

As per the question:

Diameter of the cylinder, d = 9.0 cm = 0.09 m

Length of the cylinder, l = 40 cm = 1.4 m

Average Resistivity, $\rho = 5.5\ \Omega-m$

Current, I = 100 mA = 0.1 A

Now,

To calculate the potential difference between the hands:

Cross- sectional Area of the Cylinder, A = $\pi (\frac{d}{2})^{2} = 6.36\times 10^{- 3}\ m^{2}$

Resistivity is given by:

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

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

$R = 5.5\times \frac{1.4}{6.36\times 10^{- 3}} = 1210.69\ \Omega$

Now, using Ohm's Law:

V = IR

$V = 0.1\times 1210.69 = 121.069\ V$