A person with body resistance between his hands of 11 k accidentally grasps the

Question

A person with body resistance between his hands of 11 k accidentally grasps the terminals of a 16-kV power supply.

A) If the internal resistance of the power supply is 2500 , what is the current through the person's body? Express your answer using two significant figures.

B) What is the power dissipated in his body? Express your answer using two significant figures.

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=1.00mA or less? Express your answer using two significant figures.

Explanation / Answer

Internal resistance of the power supply = r = 2500

Body resistance between hands = R = 11k =11000

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 /13500

(a) The current through the person's body = i =1.19 A

(b) The power dissipated in his body =i^2R=1.19^2*11000=15577.10 W

If Imax = 1.00 mA =0.001A
R+r=E/Imax
r =E/Imax - R
r = 16000/0.001 - 11000
r =16000000 -11000=15989000

(c) The internal resistance should be15989000 ohm for the maximum current in the above situation to be I(max) = 1.00 mA or less

The internal resistance should be1.5989000*10^7 ohm

15.99 mega ohm

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