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

The complete question is;

A person with body resistance between his hands of 10 kΩ accidentally grasps the terminals of a 16-kV power supply. What is the power dissipated in his body?

A) If the internal resistance of the power supply is 1600 Ω , what is the current through the person's body?

B) What is the power dissipated in his body?

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 I_max = 1.00mA or less?

A) I = 1.379 A

B) P = 19016.41 W

C) r = 15990000 Ω

Explanation:

A) We are given;

Internal resistance of the power supply; r = 1600 Ω

Body resistance between hands; R = 10kΩ = 10000 Ω

Power supply voltage; E =16 kV = 16000 V

Formula for the current through the person's body with internal resistance is given by;

I = E/(R + r)

Thus;

I = 16000/(10000 + 1600)

I = 1.379 A

B) Formula for power dissipated is;

P = I²R

P = 1.379² × 10000

P = 19016.41 W

C) Now, we are told that the maximum current should be I_max = 1.00mA or less. So, I_max = 0.001 A

Thus, from I = E/(R + r) and making r the subject, we have;

r = (E/I) - R

r = (16000/0.001) - 10000

r = 15990000 Ω

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