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Q94PE

Expert-verifiedFound in: Page 736

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Consider a person working in an environment where electric currents might ****pass through her body. Construct a problem in which you calculate the resistance of insulation needed to protect the person from harm. Among the things to be considered are the voltage to which the person might be exposed, likely body resistance (dry, wet, …), and acceptable currents (safe but sensed, safe and unfelt, …).**

The insulation needed to protect the student from harm had the resistance obtained as: \({R_0}{\rm{ }} = {\rm{ }}23.0{\rm{ }}\Omega \).

**Resistance is a measurement of the resistance to current flow in an electrical circuit. Resistance in ohms is denoted by the Greek letter omega**\({\rm{(\Omega )}}\)** .**

In Ohm's Law the current \(I\) through a circuit element (other than battery) can be determined by dividing the potential difference \(\Delta V\) across the circuit element by its resistance \(R\):

\(I{\rm{ }} = {\rm{ }}\frac{{\Delta V}}{R}\)

A physics students mistakenly touches a set of batteries with an emf of \(120{\rm{ }}V\). The student's hand is sweaty and has a resistance of \(1.00{\rm{ }}k\Omega \). Calculate the resistance of insulation needed to protect the student from harm knowing that the maximum harmless current is \(5.00{\rm{ }}mA\).

The emf of the battery is: \(\Delta V{\rm{ }} = {\rm{ }}120{\rm{ }}V\)

Resistance of the student's hand is:

\(\begin{align}{}R{\rm{ }} &= {\rm{ }}(1.00{\rm{ }}k\Omega )\left( {\frac{{1000{\rm{ }}\Omega }}{{1{\rm{ }}k\Omega }}} \right)\\ &= {\rm{ }}1000{\rm{ }}\Omega \end{align}\)

Maximum harmless current is:

\(\begin{align}{}I{\rm{ }} &= {\rm{ }}(5.00{\rm{ }}mA)\left( {\frac{{1{\rm{ }}A}}{{1000{\rm{ }}mA}}} \right)\\ &= {\rm{ }}5.00{\rm{ }} \times {\rm{ }}{10^{ - 3}}{\rm{ }}A\end{align}\)

harm.

Let the value of \({R_0}\) be the resistance of insulation. Applying the Ohm's law from the equation given is:

\(\Delta V{\rm{ }} = {\rm{ }}I{R_{total}}\)

The value of \({R_{total}}\) is the resistance \(R\) of the student plus the resistance \({R_0}\) of the insulation:

\(\Delta V{\rm{ }} = {\rm{ }}I(R{\rm{ }} + {\rm{ }}{R_0})\)

Rearrange and solve for the value of \({R_0}\):

\(\begin{align}{}\frac{{\Delta V}}{I}{\rm{ }} &= {\rm{ }}R{\rm{ }} + {\rm{ }}{R_0}\\{R_0}{\rm{ }} &= {\rm{ }}\frac{{\Delta V}}{I}{\rm{ }} - {\rm{ }}R\end{align}\)

Entering the values and we obtain:

\(\begin{align}{}{R_0}{\rm{ }} &= {\rm{ }}\frac{{120{\rm{ }}V}}{{5.00{\rm{ }} \times {\rm{ }}{{10}^{ - 3}}}} - 1000{\rm{ }}\Omega \\ &= {\rm{ }}2.30{\rm{ }} \times {\rm{ }}{10^4}{\rm{ }}\Omega \\ & = {\rm{ }}(2.30{\rm{ }} \times {\rm{ }}{10^4}{\rm{ }}\Omega )\left( {\frac{{1{\rm{ }}k\Omega }}{{1000{\rm{ }}\Omega }}} \right)\\ &= {\rm{ }}23.0{\rm{ }}\Omega \end{align}\)

Therefore, resistance of the insulation needed to protect the student from harm is: \({R_0}{\rm{ }} = {\rm{ }}23.0{\rm{ }}\Omega \).

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