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 15-kV power supply.

Part A

If the internal resistance of the power supply is 2200 , what is the current through the person's body?

Express your answer using two significant figures.

Part B

What is the power dissipated in his body?

Express your answer using two significant figures.

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=1.00mA or less?

Express your answer using two significant figures.

Explanation / Answer

Internal resistance of the power supply = r = 2200

Body resistance between hands = R = 11k =11000

Power supply voltage = E =15 kV=15000 V

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

The current through the person's body = i =15000 /13200

(a) The current through the person's body = i =1.1 A
_____________
(b) The power dissipated in his body =i^2R=1.4*104 W
___________________________
If Imax = 1.00 mA =0.001A

R+r=E/Imax

r =E/Imax - R

r = 15000/0.001 - 11000

r =15000000 -11000=14989000

(c) The internal resistance should be14989000 ohm for the maximum current in the above situation to be I(max) = 1.00 mA or less

The internal resistance should be14989000 ohm or 1.5*107 ohm

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