A person with body resistance between his hands of `10 K Omega` accidentally grasps the terminals of a `18 kV` power supply.
(i) If the internal resistance of the power supply is `2000 Omega`, what is the power dissipated in his body?
(ii) What is the power dissipated in his body?
(iii) 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 `1.00 mA` or less?

Explanation

To solve the problem step by step, we will break it down into three parts as per the question.

### Given Data:
- Body resistance \( R_b = 10 \, k\Omega = 10 \times 10^3 \, \Omega \)
- Power supply voltage \( V = 18 \, kV = 18 \times 10^3 \, V \)
- Internal resistance of the power supply \( R_i = 2000 \, \Omega \)

### Part (i): Calculate the Power Dissipated in the Body
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