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

Question

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

A) If the internal resistance of the power supply is 1600 , 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 Imax=1.00mA or less? Express your answer using two significant figures.

Explanation / Answer

a )

given Resistance R = 10 k = 10000

voltage V = 14 kV = 14000 V

internal resistance Power suppy r = 1600

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

i = 14000 / ( 10000 + 1600 )

i = 1.206 A

b )

the power dissipated is P = i2 R

P = 1.2062 X 10000

P = 14544.36 W

if imax =1 m A = 0.001 A

R + r = V / imax

r = V / imax - R

r = 14000 / 0.001 - 10000

r = 13990000

c )

the internal resistance be for the maximum current in the above situation to be imax= 1.00mA or less is

r = 13990000

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