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BIO 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 \(100 \mathrm{~mA}\) ? (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
The resistance between the hands is 1.28 \, \Omega, the potential difference needed for a lethal shock current of 100 \, mA is 0.128 \, V, and the power dissipated in the body with this current is 0.0128 \, W.

Step by step solution

01

Calculate Resistance

Use the formula for resistance in a cylindrical conductor, given by \(R = \frac{{\rho \cdot L}}{A}\), where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area. The cross-sectional area of a cylinder is given by \(A = \pi \cdot r^{2}\), where \(r\) is the radius. Here, \(\rho = 5.0 \, \Omega \cdot m\), \(L = 1.6 \, m\), and \(r = 0.05 \, m\) (half of the diameter). Substituting these values yields \(R = \frac{{5.0 \cdot 1.6}}{\pi \cdot 0.05^{2}} \, \Omega = 1.28\, \Omega).\)
02

Calculate Potential Difference

The potential difference can be calculated using Ohm's law, \( V = R \cdot I\), where \(V\) is the voltage, \(R\) is resistance, and \(I\) is current. Here, \(R = 1.28 \, \Omega\), and \(I = 100 \, mA = 0.1 \, A\). Substituting these values into the equation, we get \(V = 1.28 \cdot 0.1 \, V = 0.128 \, V\).
03

Calculate Power Dissipation

Use the power dissipation formula \(P = I^2 \cdot R\), where \(P\) is power, \(I\) is current, and \(R\) is resistance. We have \(\, R = 1.28 \, \Omega\) and \(\, I = 0.1 \, A\). The power dissipation is therefore \(P = (0.1)^2 \cdot 1.28\, W = 0.0128 \, W\).

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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 study of electricity. It explains how voltage, current, and resistance interact in an electric circuit. - The equation for Ohm's Law is given by \( V = I \times R \), where: - \( V \) represents the voltage across the circuit, - \( I \) is the current flowing through the circuit, and - \( R \) is the resistance of the circuit.In simpler terms, Ohm's Law tells us that the voltage (or potential difference) across a conductor is directly proportional to the current passing through it, as long as the temperature remains constant. It's essential for designing circuits and helps in predicting how changes in resistance will affect the current and voltage.When applying Ohm's Law practically, like in the human body model from our exercise, we see how even small changes in voltage can lead to significant current changes, especially when skin conductivity is altered.
Resistivity
Resistivity is a key concept when discussing materials and their ability to conduct electricity. It is fundamentally about the opposition a material poses to the flow of electric current. - The resistivity \( \rho \) of a material is - measured in ohm meters \( \Omega \cdot m \).- The formula for calculating resistance of a wire or cylinder is: \[ R = \frac{\rho \cdot L}{A} \] where: - \( R \) is the resistance, - \( \rho \) is the resistivity, - \( L \) is the length of the conductor, - \( A \) is the cross-sectional area.For instance, in the problem provided, the human body has a resistivity of \( 5.0 \Omega \cdot m \). This value implies that when the body's conductivity is considered as a cylinder, the resistance between two points can be calculated easily.The resistivity can vary significantly between materials; metals typically have low resistivity and are good conductors, whereas wood or rubber have high resistivity and are poor conductors.
Electric Circuits
Electric circuits are structured pathways designed for the flow of electricity. These circuits include components like resistors, wires, and power sources, working together to perform a function such as lighting a bulb or powering a gadget. A simple electric circuit consists of: - A voltage source (like a battery) - Conductive paths (wires) - A load or resistor (like a light bulb or a human body, as in our problem) An electric circuit allows electrons (electricity) to flow in a loop, which carries energy to and from different components. The current that flows through the circuit relies on a complete path, commonly known as a closed circuit. If there’s any break, the pathway becomes an open circuit, stopping the flow of electricity. Understanding the structure of electric circuits is crucial as it influences how electricity is controlled and used in practical applications, such as computing potential danger in scenarios similar to this exercise.
Power Dissipation
Power dissipation in an electric circuit refers to the conversion of electrical energy into other forms like heat. It's a measure of how much electrical energy is lost as heat over time in a component of the circuit.The formula to calculate power dissipation is:\[P = I^2 \times R\]Where:- \( P \) is the power- \( I \) is the current- \( R \) is the resistanceThis indicates that power dissipation increases with higher currents and also with higher resistance. It's commonly observed in resistors, heaters, and also affects electronic components which may heat up during use.In the provided exercise, we see how the body's configuration as a 'resistor' leads to a dissipation of power when a current flows through. Calculating this power loss is crucial to understanding both the potential for electrical shock and efficient energy use in circuits.

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Most popular questions from this chapter

A square, conducting, wire loop of side \(L,\) total mass \(m,\) and total resistance \(R\) initially lies in the horizontal \(x y\) -plane, with corners at \((x, y, z)=(0,0,0),(0, L, 0),(L, 0,0),\) and \((L, L, 0)\) There is a uniform, upward magnetic field \(\vec{B}=B \hat{k}\) in the space within and around the loop. The side of the loop that extends from (0,0,0) to \((L, 0,0)\) is held in place on the \(x\) -axis; the rest of the loop is free to pivot around this axis. When the loop is released, it begins to rotate due to the gravitational torque. (a) Find the net torque (magnitude and direction) that acts on the loop when it has rotated through an angle \(\phi\) from its original orientation and is rotating downward at an angular speed \(\omega .\) (b) Find the angular acceleration of the loop at the instant described in part (a). (c) Compared to the case with zero magnetic field, does it take the loop a longer or shorter time to rotate through \(90^{\circ} ?\) Explain. (d) Is mechanical energy conserved as the loop rotates downward? Explain.

BIO Electric Eels. Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to \(500 \mathrm{~V}\) and produce currents of \(80 \mathrm{~mA}\) (or even larger). A typical pulse lasts for \(10 \mathrm{~ms}\). What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?

A circular loop of wire with radius \(2.00 \mathrm{~cm}\) and resistance \(0.600 \Omega\) is in a region of a spatially uniform magnetic field \(\vec{B}\) that is perpendicular to the plane of the loop. At \(t=0\) the magnetic field has magnitude \(B_{0}=3.00 \mathrm{~T}\). The magnetic field then decreases according to the equation \(B(t)=B_{0} e^{-t / \tau},\) where \(\tau=0.500 \mathrm{~s}\). (a) What is the maximum magnitude of the current \(I\) induced in the loop? (b) What is the induced current \(I\) when \(t=1.50 \mathrm{~s} ?\)

The current in a wire varies with time according to the relationship \(I=55 \mathrm{~A}-\left(0.65 \mathrm{~A} / \mathrm{s}^{2}\right) t^{2} .\) (a) How many coulombs of charge pass a cross section of the wire in the time interval between \(t=0\) and \(t=8.0 \mathrm{~s} ?(\mathrm{~b}) \mathrm{What}\) constant current would transport the same charge in the same time interval?

An incandescent light bulb uses a coiled filament of tungsten that is \(580 \mathrm{~mm}\) long with a diameter of \(46.0 \mu \mathrm{m} .\) At \(20.0^{\circ} \mathrm{C}\) tungsten has a resistivity of \(5.25 \times 10^{-8} \Omega \cdot \mathrm{m} .\) Its temperature coefficient of resistivity is \(0.0045\left(\mathrm{C}^{\circ}\right)^{-1},\) and this remains accurate even at high temperatures. The temperature of the filament increases linearly with current, from \(20^{\circ} \mathrm{C}\) when no current flows to \(2520^{\circ} \mathrm{C}\) at 1.00 A of current. (a) What is the resistance of the light bulb at \(20^{\circ} \mathrm{C} ?\) (b) What is the current through the light bulb when the potential difference across its terminals is \(120 \mathrm{~V} ?\) (Hint: First determine the temperature as a function of the current; then use this to determine the resistance as a function of the current. Substitute this result into the equation \(V=I R\) and solve for the current \(I .\) ) (c) What is the resistance when the potential is \(120 \mathrm{~V} ?\) (d) How much energy does the light bulb dissipate in 1 min when \(120 \mathrm{~V}\) is supplied across its terminals? (e) How much energy does the light bulb dissipate in 1 min when half that voltage is supplied?

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