Problem 25.70 Part A A person with body resistance between his hands of 12 accid

Question

Problem 25.70 Part A A person with body resistance between his hands of 12 accidentally grasps the terminals of a 16- kV power supply If the internal resistance of the power supply is 2000 12, what is the current through the person's body? Express your answer using two significant figures. Submit My Answers Give Up Incorrect, Iry Again, 6 attempts remaining Part B What is the power dissipated in his body? Express your answer using two significant figures. P- 1.6x104 Submit My Answers Give Up Correct Part 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 Imax 1.00 mA or less? Express your answer using two significant figures rE Submit My Answers Give Up rovide Feedba Continue

Explanation / Answer

Given,

Internal resistance of the power supply = r = 2000 ohm

Body resistance between hands = R = 12 kohm =12000 ohm

Power supply voltage = E =16 kV=16000 V

The current through the person's body = i = E / (R+r)

The current through the person's body = i =16000 /(12000+2000) = 1.14

a] The current through the person's body = i =1.14 A

b] The power dissipated in his body = i^2R

= 1.14^2*12000

= 1.6*10^4 W

c] If Imax = 1.00 mA =0.001A

R+r = E/Imax

r =E/Imax - R

r = 16000/0.001 - 12000

r =13000000 -12000= 15988000

The internal resistance should be15988000 ohm for the maximum current in the above situation to be I(max) = 1.00 mA or less

The internal resistance should be1.5988000*10^7 ohm or 15.988 mega ohm

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