a person with body resistance between his hands of 10.0-kohms accidentally grasp
ID: 1432342 • Letter: A
Question
a person with body resistance between his hands of 10.0-kohms accidentally grasps the terminals of a 20.0-kv power supply. (do no do this!) a) draw a circuit diagram to represent the situation. b) if the internal resistance of the power supply is 2000-ohms, what is the current through his body? c) what is the power dissipated in his body? d) 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 this situation to be 1.00-mA or less? e) will this modification compromise the effectiveness of the power supply for driving low-resistance devices? explain your reasoning.
Explanation / Answer
given are
R=10,000ohm
V=20,000V
r=2000ohm
b) i = V/ (R+r)
=20,000/(10,000+2000)
I=1.66amp--b
c)P=I^2R
=1.66^2(10,000)
=27556 watt--answer b
d)If Imax = 1.00 mA =0.001A
R+r=V/Imax
r =V/Imax - R
=20,000/0.001-10,000
r=1.9*10^7 ohm--answer d
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