A person with body resistance between his hands of 13 k accidentally grasps the

Question

A person with body resistance between his hands of 13 k accidentally grasps the terminals of a 16-kV power supply. Part A If the internal resistance of the power supply is 2000 , what is the current through the person's body? Express your answer using two significant figures. I = A SubmitMy AnswersGive Up Incorrect; Try Again; 7 attempts remaining

Part B What is the power dissipated in his body? Express your answer using two significant figures. P = m W SubmitMy AnswersGive Up

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. r = m SubmitMy AnswersGive Up

Explanation / Answer

I = 16000/(13000+2000) = 1.067 A

P = I2*R = (1.067)*(1.067)*13000 = 14.8 KW

C) .001 = 16000/(13000+r)

r = 1.5987*108 ohm

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