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

Given `R = 10 k Omega and V = 18 Kv`.
(i) To find the current fl,owing through the body , we need to sum up the resistances present in the circuit and divide the voltage by it .
`I = (V) /(R + r) = ( 18 xx 10^(3))/( 10 xx 10 ^(3) + 2 xx 10^(3)) = ( 18 xx 10^(3))/( 12 xx 10^(3))`
`= (3) / (2) = 1.5 A`
(ii) Power dissipated is
` V_(1) = I^(2)R = (1.5)^(2) (10000) = 2.25 xx 10000 `
` = 22500 = 22.5 Kw`
(iii) To find the internal resistance for the safe limit of power , we can use the formula as in part `i`. The only difference here is `I` is given and `r` is to be calculated.
` R + r = (V) /(I) = ( 18 xx 10^(3))/( 1xx 10^(-3)) = 18 xx 10^(6) `
or `r = 18 xx 10^(6) - 10 xx 10^(3) = ( 18 M Omega - 10 K Omega)`
`~= 18 M Omega`