(2%) Problem 50: A careless person accidentally grasps the terminals of a 200-kV
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Question
(2%) Problem 50: A careless person accidentally grasps the terminals of a 200-kV power supply (Do NOT do this!) The resistance of the path through his body is 9.5 k2. "?33% Part (a) -? 33% Part(b) If the internal resistance of the power supply is 200 ? what is the current 1, in amperes, that passes through his body? what is the power, P, in watts, dissipated in his body? ? 33% Part (c) The power supply can be made safer by increasing its internal resistance. What should be this resistance, R the maximum current the device can supply to be 1.05 mA or less? in megohms, for Grade Summary Dedictions 0% Potential new = 11 100%Explanation / Answer
P= I^2R
I=?(P/R)
R=2000+9500
= 11500
I= ?(20000/11500)
=1.32
Power dissipate in body
= I^2R
1.32^2*9500
=16.55kV
R new
= P/I^2
20000/1.05*10^-3
=19.04 mega ohm
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