Problem 26.65 The capacitance of biological membranes is about1.0 ? F per cm2 of

Question

Problem 26.65

The capacitance of biological membranes is about1.0 ?F per cm2 of membrane area, so investigators can determine the surface area of a cell membrane by using intracellular electrodes to measure the membrane's capacitive reactance. An investigator applies a 1.0 ?Apeak current at 40 kHz to a cell and measures the peak out-of-phase voltage?that is, the component of the voltage due to the capacitive reactance of the cell membrane?to be 0.16 V .

Part A

If the investigator applies a 1.0 ?A peak current at 40 kHz to a cell with twice the membrane area of the cell noted in the passage, what will be the peak out-of-phase voltage?

Problem 26.65

The capacitance of biological membranes is about1.0 ?F per cm2 of membrane area, so investigators can determine the surface area of a cell membrane by using intracellular electrodes to measure the membrane's capacitive reactance. An investigator applies a 1.0 ?Apeak current at 40 kHz to a cell and measures the peak out-of-phase voltage?that is, the component of the voltage due to the capacitive reactance of the cell membrane?to be 0.16 V .

Part A

If the investigator applies a 1.0 ?A peak current at 40 kHz to a cell with twice the membrane area of the cell noted in the passage, what will be the peak out-of-phase voltage?

If the investigator applies a 1.0  peak current at 40  to a cell with twice the membrane area of the cell noted in the passage, what will be the peak out-of-phase voltage? 0.16 V 0.32 V 0.080 V 0.040 V

Explanation / Answer

Given

C1/A =1 uF/cm2

=>C1=A uF/cm2

So Increasing area causes

C2=2A uF/cm2

Given Voltage across capacitor is

V=I*XC =I*(1/2pi*f*C)

Since current and frequency are Constant ,so

=>V1/V2 =C2/C1

0.16/V2 =(2A/A)

V2 =0.08 Volts

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