Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.7 m

Question

Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.7 m × 1.7 m in area. A measurement of the potential difference across the inner and outer surfaces of the membrane gives a reading of 97.5 mv. Te resistivity of the membrane material is 1.30 x 107 ·m. 1.7 rm in ·-30 the membrane gives a reading of 9 (a) Determine the amount of current that flows through this portion of the membrane (b) By what factor does the current change if the side dimensions of the membrane portion is halved? The other values do not change. decrease by a factor of 8 increase by a factor of 2 increase by a factor of 4 decrease by a factor of 2 decrease by a factor of 4

Explanation / Answer

Resistance=resistivity*length/area=3.37*10^(10) ohm

current=voltage/resistance=2.89*10^(-12) A

(b) if area is decreased 4 times ,then R will 4 times

so current will decrease by factor of 4

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