A person with body resistance between his hands of 10k? accidentally grasps the

Question

A person with body resistance between his hands of 10k? accidentally grasps the terminals of a 14-kV power supply.

Part A

If the internal resistance of the power supply is 1800? , 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 = 1800 ?

Body resistance between hands = R = 10k? =10000 ?

Power supply voltage = E =14 kV=14000 V

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

The current through the person's body = i =14000 /11800

(A) The current through the person's body = i =1.186 A
_____________
(B) The power dissipated in his body =i^2R=14076.41 W
___________________________
If Imax = 1.00 mA =0.001A

R+r=E/Imax

r =E/Imax - R

r = 14000/0.001 - 10000

r =14000000 -10000=1390000 Ohm

(C) The internal resistance should be1390000 ohm for the maximum current in the above situation to be I(max) = 1.00 mA or less

The internal resistance should be1.390000*10^7 ohm or 13.9 mega ohm

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