A person with body resistance between his hands of $\mathbf{10}\mathbf{}\mathit{k}\mathit{\Omega }$accidentally grasps the terminals of a $\mathbf{14}\mathbf{-}\mathit{k}\mathit{V}$power supply. (a) If the internal resistance of the power supply is $\mathbf{2000}\mathit{\Omega }$, 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 $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathit{A}$or less?

Use the formula of current

$I=\frac{\epsilon }{R+r}$

Substitute the given values

$$\begin{gathered} I=\frac{14000V}{10000\Omega +2000\Omega } \\ I=1.17A \end{gathered}$$

Therefore, If the internal resistance of the power supply is $2000\Omega$, the current through the person’s body is 1.17 A

Use the formula of power in relation with current and resistance

$P=I^{2}R$

Substitute the values

$$\begin{gathered} P={\left(1.17A\right)}^2\left(10000\Omega \right) \\ P=13.7kW \end{gathered}$$

Therefore, the power dissipated through the person’s body is$13.7kW$

Use the formula of current used above and transform it accordingly

$r=\frac{\epsilon -IR}{I}$

Substitute the values

$$\begin{gathered} r=\frac{14000V-\left(0.001A\right)\left(10000\Omega \right)}{0.001A} \\ r=14M\Omega \end{gathered}$$

Therefore, the internal resistance should be $14M\Omega$for the maximum current in the above situation to be 1.00 mA or less