Question: BIO The average bulk resistivity of the human body (apart from surface resistance of the skin) is about . The conducting path between the hands can be represented approximately as a cylinder long and in diameter. The skin resistance can be made negligible by soaking the hands in salt water. (a) What is the resistance between the hands if the skin resistance is negligible? (b) What potential difference between the hands is needed for a lethal shock current of ? (Note that your result shows that small potential differences produce dangerous currents when the skin is damp.) (c) With the current in part (b), what power is dissipated in the body?

Consider the expression for the resistance is:

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

Here, $\rho$is resistivity $\Omega ·\text{m}$, L is length in A and is area in ${\text{m}}^2$.

Consider the formula for the power dissipated in the circuit.

$P=I^{2}R$ ….. (1)

Consider the given values of resistivity, length and the diameter.

$$\begin{gathered} \rho =5.0\Omega ·\text{m} \\ L=1.6\text{m} \\ d=0.10\text{m} \end{gathered}$$

Substitute the values in resistance formula:

$$\begin{gathered} R=\frac{\rho L}{A} \\ R=\frac{\rho L}{\pi {\left(\frac{d}{2}\right)}^2} \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} R=\frac{\left(5.0\Omega ·\text{m}\right)\left(1.6\text{m}\right)}{\pi {\left(\frac{0.10}{2}\right)}^2} \\ =1019\Omega \end{gathered}$$

Hence, the resistance between the hands if the skin resistance is negligible is $1019\Omega$.

(b)

Consider the value of the current.

$I=100\text{mA}$

Consider the expression value of the voltage from the Ohm’s law.

$$\begin{gathered} V=\left(100\text{mA}\right)\left(1019\Omega \right) \\ =102\text{V} \end{gathered}$$

Hence, the potential difference between the hands is needed for a lethal shock current of $100\text{mA}$is 102 V .

(c)

Consider the value of the current.

$I=100\text{mA}$

Substitute the values in the equation (1) to determine the power.

$$\begin{gathered} P={\left(100\text{mA}\right)}^2\left(1019\Omega \right) \\ =10.19\text{W} \end{gathered}$$

Hence, power is dissipated in the body.