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 =1.136 A
(b) The power dissipated in his body =i^2R=1.136^2*11000=12500 W
If Imax = 1.00 mA =0.001A
R+r=E/Imax
r =E/Imax - R
r = 15000/0.001 - 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 be1.4989000*10^7 ohm or 14.989 mega ohm

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