91Ó°ÊÓ

Conductive tissues length $=40cm$

Conductive tissues diameter$=12cm$

$Resistivit{y}_1=12\mathrm{Î\copyright }m$

$Resistivit{y}_2=25\mathrm{Î\copyright }m$

Person's leg musclerole="math" localid="1649150503366" $=82\%$

Person's leg fat $=18\%$

Voltage $=1.5V$

Let us recall that according to Ohm's law, the current will be

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

The resistance will be provided as a function of the length, area - which is a circle - of the cross-section, and resistivity as

$R=\mathrm{ÒÏ}\frac{L}{A}=\mathrm{ÒÏ}\frac{4L}{\mathrm{Ï€}{D}^2}$

What we will have is two resistors, connected in parallel. Therefore the equivalent resistance will be

$R_{eq}=\frac{R_1{R}_2}{R_1+R_2}$

While we can continue with a completely parametric explanation, it won't be as likely the efforts are not worth the result.

The parameters that change in the resistances are the resistivities and a stable added by the area since we are given the percentage of muscle and fat in the leg.

That means that we can calculate the resistances to be

$R_1=\left(13\mathrm{Î\copyright }m×\frac{4×0.4m}{0.82×\mathrm{Ï€}×0.12m^2}\right)=561\mathrm{Î\copyright }$

$R_2=\left(25\mathrm{Î\copyright }m×\frac{4×0.4m}{0.18×\mathrm{Ï€}×0.12m^2}\right)=4912\mathrm{Î\copyright }.$

With this understanding, we currently proceed as we would in a normal process with two resistors in parallel. The equivalent resistance will be

$R_{eq}=\left(\frac{561\mathrm{Î\copyright }×4912\mathrm{Î\copyright }}{561\mathrm{Î\copyright }+4912\mathrm{Î\copyright }}\right)=\mathit{503}\mathit{.}\mathit{5}\mathrm{Î\copyright }$

Now, knowing the potential difference applied, we get

$I=\frac{V}{R_{eq}}=\left(\frac{1.5V}{503.5\mathrm{Î\copyright }}\right)=2.98mA$

The potential difference is applied between the person’s hip and knee found to be $2.98mA$