Question

The conducting path between the right hand and the left hand can be modeled as a 10-cm-diameter, 160-cm-long cylinder. The average resistivity of the interior of the human body is 5.0 Ω m. 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 mA across the chest? Your result shows that even small potential differences can produce dangerous currents when the skin is wet.

Short answer

The potential difference required for lethal shock is $101.86\mathrm{V}$.

Step-by-step solution

Given Information

Given in the question,

Set $d=10\mathrm{cm},l=160\mathrm{cm}$and $\rho =5.0\Omega \mathrm{m}$, and $I=100\mathrm{mA}$.

Explanation

The area of the cross-section of the cylinder that is used to model the route between your hands is

$A=\frac{1}{4}\pi d^2$

The resistance of this cylinder is

$R=\rho \frac{l}{A}=\rho \frac{4l}{\pi d^2}$

The current and the potential difference are related by Ohm's law

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

Calculate the resistivity,

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

The expression for the potential difference,

$\mathrm{\Delta }V=\frac{4\rho IL}{\pi d^2}$

The diameter of the arms is $10\mathrm{cm}$.

Convert units of diameter from $cm$to $m,$

$d=10\mathrm{cm}$

$=(10\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

$=0.1\mathrm{m}$

The length of arm is $160\mathrm{cm}$.

Convert units of length from $cm$to $m,$

$L=160\mathrm{cm}$

$=(160\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

$=1.6\mathrm{m}$

Current across the chest is $100\mathrm{mA}$.

Convert units of current from $\mathrm{mA}$to $\mathrm{A}$.

$I=100\mathrm{mA}$

$=(100\mathrm{mA})\left(\frac{1\mathrm{A}}{1000\mathrm{mA}}\right)$

$=0.1\mathrm{A}$

Explanation

Substitute $\frac{4\rho L}{\pi d^2}$for $R$in the equation $\Delta V=IR$and solve for $∆V,$

$\Delta V=I\left(\frac{4\rho L}{\pi d^2}\right)$

$=\frac{4\rho IL}{\pi d^2}$

Substitute the values,

role="math" localid="1649150118248" $\mathrm{\Delta }V=\left(\frac{4(5.0\mathrm{\Omega }\cdot \mathrm{m})\left(0.100\mathrm{A}\right)\left(1.60\mathrm{m}\right)}{\pi (0.10\mathrm{m}{)}^2}\right)$

$=101.86\mathrm{V}$

Final Answer

Therefore, the potential difference required for lethal shock is$101.86V$