Safe Resista
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If the power supply is to be made safe by increasing its internal resistance, what should the internal resista
A person with body resistance between his hands of 11 kohms accidentally grasps the terminals of a 16kV- power supply.
A) If the internal resistance of the power supply is 1700ohms , 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 or less?
Express your answer using two significant figures
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Internal resistance of the power supply = r = 1700 Ω
Body resistance between hands = R = 11kΩ =11000 Ω
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 /12700
(a) The current through the person's body = i =1.2958 A
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(b) The power dissipated in his body =i^2R=17459.23 W
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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 be1.5989*10^7 ohm or 15.989 mega ohm
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