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The average bulk resistivity of the human body (apart from surface resistance of the skin) is about 5.0\(\Omega \cdot \mathrm{m}\) . The conducting path between the hands can be represented approximately as a cylinder 1.6 \(\mathrm{m}\) long and 0.10 \(\mathrm{m}\) in diameter. The skin resistance can be made negligible by soaking the hands in salt water. (a) What is the resistance between the hands if the skin resistance is negligible? (b) What potential difference between the hands is needed for a lethal shock current of 100 \(\mathrm{mA}\) A? (Note that your result shows that small potential differences produce dangerous currents when the skin is damp.) (c) With the current in part (b), what power is dissipated in the body?

Short Answer

Expert verified
(a) 1019 Ω, (b) 101.9 V, (c) 10.19 W.

Step by step solution

01

Determine the cross-sectional area of the cylinder

The cylinder has a diameter of 0.10 m, so its radius is \( r = \frac{0.10}{2} = 0.05 \) m. The cross-sectional area \( A \) of the cylinder is given by the area formula of a circle: \( A = \pi r^2 \), thus,\[A = \pi (0.05)^2 = 0.00785 \, \text{m}^2.\]
02

Use the formula for resistance to find the resistance between the hands

The resistance \( R \) of a cylindrical object can be calculated using the formula \( R = \rho \frac{L}{A} \), where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. Given \( \rho = 5.0 \, \Omega \cdot \text{m} \), \( L = 1.6 \, \text{m} \), and \( A = 0.00785 \, \text{m}^2 \):\[R = 5.0 \times \frac{1.6}{0.00785} = 1019 \, \Omega.\]
03

Calculate the potential difference for a lethal shock current

Ohm’s Law states that \( V = I \times R \), where \( V \) is the potential difference, \( I \) is the current, and \( R \) is the resistance. For a current of 100 mA (or 0.1 A):\[V = 0.1 \, \text{A} \times 1019 \, \Omega = 101.9 \, \text{V}.\]
04

Find the power dissipated in the body

The power \( P \) dissipated can be calculated using the formula \( P = I^2 \times R \) or \( P = V \times I \). Here, using the current:\[P = (0.1)^2 \times 1019 = 10.19 \, \text{W}.\] This calculation confirms that a similar result would be obtained if using the potential difference: \( P = 101.9 \times 0.1 = 10.19 \, \text{W} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
Ohm's Law is a fundamental principle in the world of electricity and circuits. It relates the voltage (potential difference), current, and resistance in an electric circuit, and is usually expressed with the equation \( V = I \times R \). Here, \( V \) represents the voltage, \( I \) is the current, and \( R \) is the resistance.

This simple equation can help predict how a circuit will respond under various conditions. If you increase the voltage while keeping resistance constant, the current increases. Similarly, for a fixed voltage, increasing the resistance will reduce the current.

In practical terms, Ohm's Law helps in designing circuits. By knowing any two of the three quantities, you can calculate the third. This is crucial for ensuring devices work safely and efficiently, and helps engineers create suitable electrical pathways to prevent damage or malfunctions.
Resistivity
Resistivity is a property of materials that quantifies how strongly a given material opposes the flow of electric current. It's represented by the symbol \( \rho \) and is measured in ohm-meters (\( \Omega \cdot \text{m} \)).

Different materials have different resistivities. Metals, for example, typically have low resistivity, making them good conductors of electricity. In contrast, rubber and glass have high resistivity, making them good insulators.

The resistivity of materials is important for determining the resistance of an object, using the formula \( R = \rho \frac{L}{A} \), where \( L \) is the length and \( A \) is the cross-sectional area. The longer and thinner a material is, the greater its resistance will be for a given resistivity.

Understanding resistivity helps in selecting materials for various electrical and electronic applications. By choosing the appropriate material, engineers can control the flow of electric current and optimize circuit performance.
Power Dissipation
Power dissipation in an electrical circuit refers to the conversion of electric energy into other forms, like heat. It is often a critical factor when designing circuits and electrical devices, as excessive power dissipation can lead to overheating and damage.

Power dissipation is calculated using the formula \( P = I^2 \times R \) or \( P = V \times I \). Both equations reflect how power changes depending on current, voltage, and resistance. Using \( I^2 \times R \), we observe that doubling the current will increase the power by a factor of four, assuming resistance remains unchanged.

Design considerations for power dissipation include ensuring enough cooling is provided, such as using heat sinks and fans, to manage the heat generated. Safe levels of power dissipation ensure long-term reliability and efficiency of the system.

Knowing how to calculate and manage power dissipation aids in building robust and efficient electrical systems that are less prone to failure.
Current and Voltage
Current and voltage are two essential concepts in understanding electricity and circuits. Current, measured in amperes (A), is the flow of electric charge through a conductor. Voltage, measured in volts (V), is the potential energy per charge.

In simple terms, think of current like the flow of water in a pipe, where voltage is the pressure that pushes the water to move. More voltage means more potential energy to move the electrons, resulting in more current if the circuit allows it. Switching the analogy back to circuits, a higher current flows when the voltage is high and resistance is low.

Circuits are often categorized by how they manage current and voltage. For instance, a series circuit passes the same current through all components, while in a parallel circuit, the components share the same voltage but each may have different currents.

Understanding the relationship between current and voltage is crucial for circuit design and analysis. It helps in selecting the right components to achieve desired electrical behavior and safety.

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Most popular questions from this chapter

The potential difference between points in a wire 75.0 \(\mathrm{cm}\) apart is 0.938 \(\mathrm{V}\) when the current density is \(4.40 \times 10^{7} \mathrm{A} / \mathrm{m}^{2}\) . What are (a) the magnitude of \(\overrightarrow{\boldsymbol{E}}\) in the wire and \((\mathrm{b})\) the resistivity of the material of which the wire is made?

A \(5.00-\mathrm{A}\) current runs through a 12 -gauge copper wire (diameter 2.05 \(\mathrm{mm} )\) and through a light bulb. Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

The potential difference across the terminals of a battery is 8.4 \(\mathrm{V}\) when there is a current of 1.50 \(\mathrm{A}\) in the battery from the negative to the positive terminal. When the current is 3.50 \(\mathrm{A}\) in the reverse direction, the potential difference becomes 9.4 \(\mathrm{V}\) . (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

The resistivity of a semiconductor can be modified by adding different amounts of impuritics. A rod of semiconducting material of length \(L\) and cross- sectional area \(A\) lies along the \(x\) -axis between \(x=0\) and \(x=L\) . The material obeys Ohm's law, and its resistivity varies along the rod according to \(\rho(x)=\) \(\rho_{0} \exp (-x / L) .\) The end of the rod at \(x=0\) is at a potential \(V_{0}\) greater than the end at \(x=L\) (a) Find the total resistance of the rod and the current in the rod. (b) Find the electric-field magnitude \(E(x)\) in therod as a function of \(x\) (c) Find the electric potential \(V(x)\) in the rod as a function of \(x\) . (d) Graph the functions \(\rho(x) . E(x)\) . and \(V(x)\) for values of \(x\) between \(x=0\) and \(x=L\) .

The following measurements of current and potential difference were made on a resistor constructed of Nichrome wire: $$ \begin{array}{l|llll}{\mathbf{I}(\mathbf{A})} & {0.50} & {1.00} & {2.00} & {4.00} \\ {V_{a b}(\mathbf{v})} & {1.94} & {3.88} & {7.76} & {15.52}\end{array} $$ (a) Graph \(V_{a b}\) as a function of \(I .\) (b) Does Nichrome obey Ohm's law? How can you tell? (c) What is the resistance of the resistor in ohms?

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