Question

3. In some countries AC outlets near bathtubs are restricted to a maximum of 25 V to minimize the chance of dangerous shocks while bathing. A man is in the tub; the lower end of his torso is well grounded, and the skin resistance of his wet, soapy hands is negligible. He reaches out and accidentally touches a live electric wire. What voltage on the wire would produce a dangerous 100 mA current

Answer 1

Answer:

The answer is below

Explanation:

The average resistivity of the human body (apart from surface resistance of the skin) is about 5.0 Ωm. The conducting path between the right and left hands can be approximated as cylinder 1.6 m long and 0.10 m in diameter. The skin resistance can be made negligible by soaking the hands in salt water. (a) What is the resistance between the hands if the skin resistance is negligible? (b) if skin resistance is negligible, what potential difference between the hands is needed for lethal shock current 100 mA?

Solution:

The resistance (R) of a material is given by the formula:

R = ρL / A

where L is the length of the conductor, ρ is resistivity and A is the cross sectional area.

a) R = ρL / A

ρ = 5.0 Ωm, L = 1.6 m, A = π(diameter²) / 4 = π(0.1²)/4 = 0.00785 m²

R = 5(1.6) / 0.00785 = 1018.6 ohm

b) To produce 100 mA, we need to use ohms law:

I = 100 mA = 0.1 A, R = 1018.6 ohm

V = IR = 0.1(1018.6)

V = 101.86 V