Consider a portion of a cell membrane that has a thickness of 7.50nm and 1.3 mic

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

Explanation / Answer

A =(1.3*10^-6)^2 =1.69*10^-12 m^2

R=pL/A =(1.3*10^7)(7.5*10^-9)/(1.69*10^-12)

R=5.77*10^10 ohms

I=V/R =92.2*10^-3/5.77*10^10

I=1.60*10^-12 A

b)

S=So/2=1.3/2 =0.65um

A=(0.65*10^-6)^2=4.225*10^-13 m^2

R=1.3*10^7*7.5*10^-9/(4.225*10^-13)

R=2.31*10^11 ohms

I =92.2*10^-3/2.31*10^11

I=4*10^-13 A

I1/I2 =1/4

I2 =(1/4)I1

so decreases by a factor of 4

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