A person with body resistance between his hands of 10KÙ accidently grasps the terminal of a 14-KV power supply.

Question

(a) If the internal resistance of the power supply is 2000Ù, what is the current through the person’s body?
(b) What is the power dissipated in his body?
(c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00mA or less?

Answer 1

(a) I = V/R
where R = Rb + Ri
= (10 + 2)*10^3 = 12,000 ohms

Rb = body resistance
Ri = power supply internal resistance

(b) I^2*Rb

(c) Require that V/(Rb + Ri) < 10^-3 A
Solve for Ri