Suggested languages for you:

Americas

Europe

Q94PE

Expert-verified
Found in: Page 736

### College Physics (Urone)

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: $${R_0}{\rm{ }} = {\rm{ }}23.0{\rm{ }}\Omega$$.

See the step by step solution

## Step 1: Define Resistance

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

## Step 2: Concepts and Principles

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 $$R$$:

$$I{\rm{ }} = {\rm{ }}\frac{{\Delta V}}{R}$$

## Step 3: The Constructed Problem

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 $$5.00{\rm{ }}mA$$.

## Step 4: The given data and required data

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.

## Step 5: Evaluating the resistance of the insulation needed to protect the student from 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: $${R_0}{\rm{ }} = {\rm{ }}23.0{\rm{ }}\Omega$$.