Question

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?

Answer 1

To calculate and solve the problem it is necessary to apply the concepts related to resistance and resistivity.

The equation that is responsible for relating the two variables is:

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

Where,

R= Resistance of the conductor

$\rho =$Resistivity of the conductor material

L = Length

A = Cross-sectional area of conductor

With the previous values the area of the muscle (Real Muscle-82%)is,

$A_m = (0.82)\pi r^2 = (0.82)\pi (12/2*10^{-2})^2$

$A_m = 9.274*10^{-3}m^2$

Using the equation from Resistance we have that at the muscle the value is:

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

$R_m = \frac{13*(0.4)}{9.274*10^{-3}}$

$R_m = 560.70\Omega$

At the same time we can make the same process to calculate the resistance of the fat, then

$A_m = (0.18)\pi r^2 = (0.18)\pi (12/2*10^{-2})^2$

$A_m = 2.0357*10^{-3}m^2$

The resistance of the fat would be,

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

$R_f = \frac{25*(0.4)}{2.0357*10^{-3}}$

$R_f = 4912.3\Omega$

Then the total resistance in a set as the previously writen, i.e, in parallel is:

$R=\frac{R_mR_f}{R_m+R_f}$

$R = \frac{(560.70)(4912.3)}{4912.3+560.70}$

$R = 502.62\Omega$

We can here apply Ohm's law, then

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

$I = \frac{1.5}{502.62}$

$I = 2.984*10^{-3}A$

$I = 2.984mA$

Answer 2

The measure of current in the body of the person, due to the upper leg is 0.0027 A.

What is the Ohm's law?

Ohm's law states that for a flowing current the potential difference of the circuit is directly proportional to the current flowing in it. Thus,

$V\propto I$

Here, (V) is the potential difference and (I) is the current It can be written as,

$V=IR$

Here, (R) is the resistance of the circuit.

The resistivity of muscle and fat are 13 Ω m and 25 Ω m, respectively. One person’s upper leg is 82% muscle, 18% fat.

Thus, the resistivity of the upper leg, can be given as,

$\rho=0.82\times0.13+0.18\times0.25\\\rho=15.16 \rm ohm-m$

The formula of resistance in terms of resistivity can be given as,

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

As the conductive tissues of the upper leg can be modeled as a 40-cm-long, 12-cm-diameter cylinder of muscle and fat. Thus, put the values in the above formula as,

$R=\dfrac{15.16\times0.40}{\dfrac{\pi(0.12)^2}{4}}\\R=551.27 \rm ohm$

As the potential difference applied between the person’s hip and knee is 1.5 V.Now. Thus, put the values in the ohm's law as,

$1.5=I\times551.27\\I=0.0027 \rm A$

Thus the measure of current in the body of the person, due to the upper leg is 0.0027 A.

Learn more about the Ohm's law here;

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