Question

Suppose that the resistance between the walls of a biological cell is 5.3 × 10⁹ Ω. (a) What is the current when the potential difference between the walls is 79 mV? (b) If the current is composed of Na⁺ ions (q = +e), how many such ions flow in 0.59 s?

Answer 1

given

resistance between the walls of a biological cell is R = 5.3 × 109 Ω

and

the potential differnce between the walls is v = 79 mV

the current , potential and resistance relation by the Ohm's law is

a)

V = I*R

I = V/R

I = (79*10^-3)/(5.3*10^9) A

I = 14.905660377358*10^-12 A

I = 14.9057 pA

b)

the definition of a current is the number of charges crossing the cross section are per unit time that is  

I = Q/t

I = n*q/t

n = I*t/q

here q = 1.6*10^-19 C

n = 14.9057*10^-12*0.59/(1.6*10^-19) ions

n = 54964769 ions can flow