Question

Suppose that the resistance between the walls of a biological cell is 4.2 × 109 Ω. (a) What is the current when the potential difference between the walls is 75 mV? (b) If the current is composed of Na+ ions (q = +e), how many such ions flow in 0.74 s?

Answer 1

(a) Current will be
`17.857* 10^(-12)A`

(b) Number of ions will be
`8.258* 10^6`

Step-by-step explanation:

We have given that resistance of the biological cell
`R=4.2* 10^9ohm`

(a) We have given potential difference of 75 mV

So
`V=75* 10^(-3)volt`

From ohm's law we know that current is given by

`i=(V)/(R)=(75* 10^(-3))/(4.2* 10^9)=17.857* 10^(-12)A`

(b) We have given time t = 0.74 sec

We have to find the charge

We know that charge is given by Q = it, here i is current and t is time

So charge will be
`Q=17.857* 10^(-12)* 0.74=13.214* 10^(-12)C`

So number of ions will be
`n=(13.214* 10^(-12))/(1.6* 10^(-19))=8.258* 10^6`