If one does not take into account the surface resistance of the skin, the averag

Question

If one does not take into account the surface resistance of the skin, the average resistivity of the human body is about 5m . The conducting path between the hands stretched out can be approximated as a 1.6m long cylinder with the diameter of 0.1m. (a) What is the resistance of this conductor? (As a side note, the resistance of the skin is considerably larger for dry skin than for wet skin, so one could imagine a worst-case scenario in which the skin resistance was lowered to a negligible value by wetting the skin with salt water. This, of course, should never be done when operating electrical devices! The resistance of very dry skin can be as high as 500k, whereas that of wet skin can be as low as 1k). (a) What is the resistance of this conductor? (b) What voltage between the hands would correspond to a lethal shock current of 100mA? (c) What power is dissipated through the body? (d) If the resistance between the hands is increased to 10k and a person accidentally grasps the terminals of a 14kV power supply with an internal resistance of 2k, what is the current through the person’s body? Draw a circuit representing the situation prior to writing down the appropriate equation and solving for the current. What is the power dissipated in the person’s body? (e) Could the power supply be made safe and how would that be accomplished? One could require a maximum unharmful current of 1mA.

Explanation / Answer

a) R = rho * L / A

L = 1.6 m
A = pi d^2 / 4 = pi * 0.1^2 / 4 = 7.85 x 10^-3 m^2

rho = 5 ohm . m

R = 1018.6 ohm

b) V = I R = 0.100 x 1018.6 =101.86 Volt

c) P = I^2 R = 0.1^2 ( 1018.6) = 10.2 W

d) Req = 10 +2 = 12kohm

I = V/Req = 14kV/12kohm = 1.17 A

P = I^2 R = 1.17^2 x 10 x 10^3 = 13611.11 W

e) by adding large resistor in series.

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