The conductive tissues of the upper leg can be modeled as a 40-cm-long, 12-cm-diameter cylinder of muscle and fat. The resistivities of muscle and fat are 13 Ω m and 25 Ω m, respectively. One person’s upper leg is 82% muscle, 18% fat. What current is measured if a 1.5 V potential difference is applied between the person’s hip and knee?

Conductive tissues length $=40cm$

Conductive tissues diameter$=12cm$

$Resistivit{y}_1=12\Omega m$

$Resistivit{y}_2=25\Omega m$

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

$R=\rho \frac{L}{A}=\rho \frac{4L}{\pi D^2}$

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

$R_1=\left(13\Omega m\times \frac{4\times 0.4m}{0.82\times \pi \times 0.12m^2}\right)=561\Omega$

$R_2=\left(25\Omega m\times \frac{4\times 0.4m}{0.18\times \pi \times 0.12m^2}\right)=4912\Omega .$

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

$R_{eq}=\left(\frac{561\Omega \times 4912\Omega }{561\Omega +4912\Omega }\right)=\mathit{503}\mathit{.}\mathit{5}\Omega$

Now, knowing the potential difference applied, we get

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

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