Question

Suppose that the resistance between the walls of a biological cell is 6.8 × 109 ω. (a) what is the current when the potential difference between the walls is 69 mv? (b) if the current is composed of na+ ions (q = +e), how many such ions flow in 0.86 s?

Answer 1

(a) We can find the current flowing between the walls by using Ohm's law:
$I= \frac{\Delta V}{R}$
where $\Delta V=69 mV=0.069 V$ is the potential difference and $R=6.8\cdot 10^9 \Omega$ is the resistance. Substituting these values, we get
$I=1.01 \cdot 10^{-11} A$

(b) The total charge flowing between the walls is the product between the current and the time interval:
$Q=I \Delta t$
The problem says $\Delta t=0.86 s$, so the total charge is
$Q=(1.01\cdot 10^{-11} A)(0.86 s)=8.73 \cdot 10^{-12} C$

The current consists of Na+ ions, each of them having a charge of $e=1.6 \cdot 10^{-19} C$. To find the number of ions flowing, we can simply divide the total charge by the charge of a single ion:
$N= \frac{Q}{e} = \frac{8.73 \cdot 10^{-12}C}{1.6 \cdot 10^{-19}C} = 5.45 \cdot 10^7 ions$