Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.8 Co

Question

Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.8

Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.8 A mum A? 1.8 A mum in area. A measurement of the potential difference across the inner and outer surfaces of the membrane gives a reading of 88.5 mV. The resistivity of the membrane material is 1.30 x 107 ? Am. (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 doubled? The other values do not change.

Explanation / Answer

V = 88.5 mV = 0.0885 V

resistivity , r = 1.3*10^7 ohm.m

d = 7.5 nm = 7.5*10^-9 m

A = 1.8um * 1.8um = 1.8^2*10^-12 m2 = 3.24*10^-12 m2

a)

So, resistance, R = r*d/A

So, R = 1.3*10^7*7.5*10^-9/(3.24*10^-12)

So, R = 3*10^10 ohm

So, Current, I = V/R = 0.0885/(3*10^10) = 2.95*10^-12 A <-------------answer

b)

If side dimensions are doubled the area becomes 4 times. and the thickness 2 times

So, R = Ro*2 times/4 times = 2*Ro <----------- 2 times the resistance

So, current changes by V/(2Ro) = 0.5*times the original <------- current becomes half

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