Question

Consider a portion of a cell membrane that has a thickness of 7.50nm and 1.3 micrometers x 1.3 micrometers in area. A measurement of the potential difference across the inner and outer surfaces of the membrane gives a reading of 92.2mV. The resistivity of the membrane material is 1.30 x 10^7 ohms*m



PLEASE SHOW WORK!

a) Determine the amount of current that flows through this portion of the membrane

Answer: _____A



b) By what factor does the current change if the side dimensions of the membrane portion is halved? The other values do no change

increase by factor of 2

decrease by factor of 8

decrease by factor of 2

decrease by a factor of 4

increase by factor of 4

Answer 1

Answer:

(a). The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). The factor of the current is increase by factor of 2.

(1) is correct option

Explanation:

Given that,

Thickness = 7.50 nm

Area $A=(1.3\times1.3\times10^{-6})^2$

Potential difference = 92.2 mV

Resistivity of the material $\rho=1.30\times10^{7}\ \ohm$

We need to calculate the resistance

Using formula of resistivity

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

Put the value into the formula

$R=\dfrac{1.30\times10^{7}\times7.50\times10^{-9}}{(1.3\times1.3\times10^{-6})^2}$

$R=3.413\times10^{10}\ \Omega$

(a). We need to calculate the amount of current that flows through this portion of the membrane

Using Ohm's law

$V=IR$

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

Put the value into the formula

$I=\dfrac{92.2\times10^{-3}}{3.413\times10^{10}}$

$I=2.70\times10^{-12}\ A$

The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). If the side dimensions of the membrane portion is halved

We need to calculate the new resistance

Using formula of resistivity

$R'=\dfrac{\rho \dfrac{l}{2}}{A}$

Put the value into the formula

$R'=\dfrac{1.30\times10^{7}\times\dfrac{7.50\times10^{-9}}{2}}{(1.3\times1.3\times10^{-6})^2}$

$R'=1.706\times10^{10}\ \Omega$

We need to calculate the new current

Using Ohm's law

$V=I'R'$

$I'=\dfrac{V}{R'}$

Put the value into the formula

$I'=\dfrac{92.2\times10^{-3}}{1.706\times10^{10}}$

$I'=5.404\times10^{-12}\ A$

We need to calculate the factor

$\dfrac{I'}{I}=\dfrac{5.404\times10^{-12}}{2.70\times10^{-12}}$

$I'=2I$

The factor of the current is increase by factor of 2.

Hence,(a). The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). The factor of the current is increase by factor of 2.