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

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

(a). The amount of current that flows through this portion of the membrane is

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

(1) is correct option

Explanation:

Given that,

Thickness = 7.50 nm

Area

Potential difference = 92.2 mV

Resistivity of the material

We need to calculate the resistance

Using formula of resistivity

Put the value into the formula

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

Using Ohm's law

Put the value into the formula

The amount of current that flows through this portion of the membrane is

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

We need to calculate the new resistance

Using formula of resistivity

Put the value into the formula

We need to calculate the new current

Using Ohm's law

Put the value into the formula

We need to calculate the factor

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

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

the answer would be 32*f

no. dot product   =0 when vecs at right angles