Suppose that the resistance between the walls of a biological cell is 7.0 × 109 Ω. (a) What is the current when the potential difference between the walls is 80 mV? (b) If the current is composed of Na+ ions (q = +e), how many such ions flow in 0.85 s?

Explanation:

Using Ohm's law

V ( voltage) = I (current A) × Resistance R in ohms

R = 7.0 × 10⁹Ω

V = 80 mV = 80 / 1000 = 0.08 V

0.08 V = I × 7.0 × 10⁹Ω

a) I = 0.08 V / 7.0 × 10⁹Ω = 1.142857 × 10 ⁻¹¹ A

b) quantity of charge = I × t = 1.142857 × 10 ⁻¹¹ A × 0.85 s = 9.7142857 × 10⁻¹² C

number of Na⁺ ions ( q = +e) = 9.7142857 × 10⁻¹² C / 1.6 × 10⁻¹⁹ C = 60714285.714 Na⁺ ions

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