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The average bulk resistivity of the human body (apart from surface resistance of the skin) is about 5.0\(\Omega\) \(\cdot\) m. The conducting path between the hands can be represented approximately as a cylinder 1.6 m long and 0.10 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 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
(a) 1018.34 Ω, (b) 101.834 V, (c) 10.1834 W

Step by step solution

01

Calculate the Cross-Sectional Area

The cross-sectional area of a cylinder with diameter \(d = 0.10\, m\) is given by the formula: \[ A = \pi \left( \frac{d}{2} \right)^2 \]First, calculate the radius: \( \frac{d}{2} = 0.05\, m \).Then, substitute the radius into the area formula: \[ A = \pi (0.05)^2 = \pi \times 0.0025 \, m^2 \approx 0.00785 \, m^2 \]
02

Calculate the Bulk Resistance

The resistance \(R\) of a cylindrical conductor is calculated with the formula: \[ R = \frac{\rho \cdot L}{A} \]where \(\rho = 5.0 \, \Omega \cdot m\), \(L = 1.6\, m\), and \(A \approx 0.00785 \, m^2\). Substitute these values:\[ R = \frac{5.0 \times 1.6}{0.00785} \approx 1018.34 \, \Omega \]
03

Calculate Potential Difference for Lethal Current

To find the potential difference for a current \(I = 100\, mA = 0.1\, A\), we use Ohm's Law: \[ V = I \cdot R \]Substitute \(R = 1018.34\, \Omega\) and \(I = 0.1\, A\):\[ V = 0.1 \times 1018.34 = 101.834 \, V \]
04

Calculate Power Dissipated in the Body

The power \(P\) dissipated is given by the formula: \[ P = I^2 \cdot R \]Substitute \(I = 0.1\, A\) and \(R = 1018.34\, \Omega\):\[ P = (0.1)^2 \times 1018.34 = 0.01 \times 1018.34 = 10.1834 \, W \]

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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 cornerstone concept in the study of electrical circuits. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant. Mathematically, it is expressed as:
  • \( V = I \cdot R \)
Here, \( V \) is the potential difference (voltage) in volts, \( I \) is the current in amperes, and \( R \) is the resistance in ohms.
This fundamental principle helps us understand how electrical circuits operate, making it a crucial element in the analysis of electrical systems.
In the context of our exercise, Ohm's Law was used to determine the voltage necessary to create a specific current flow through the human body, represented as a cylindrical conductor. This shows how even a small voltage can result in a dangerous current if the resistance is sufficiently low.
Bulk resistivity
Bulk resistivity is a material property that quantifies how strongly a material opposes the flow of electric current. It's essentially the resistance of a material regardless of its shape or size. The formula to determine resistance based on bulk resistivity is:
  • \( R = \frac{\rho \cdot L}{A} \)
Where \( \rho \) is the bulk resistivity in ohm-meters, \( L \) is the length of the material in meters, and \( A \) is the cross-sectional area in square meters.
Bulk resistivity is particularly useful when you need to calculate the resistance of a material configured into shapes like wires or cylinders, as seen in our exercise. By knowing the resistivity of the human body, and the approximate dimensions of the conducting path, we could calculate the body's resistance. This step not only aids in academic exercises but also has practical implications in safety standards, where understanding human-body resistivity can help prevent electrical accidents.
Power dissipation
Power dissipation refers to the process in which electrical energy is converted into heat energy in a resistive component. The power dissipated in a resistor can be calculated from the formula:
  • \( P = I^2 \cdot R \)
Where \( P \) is power in watts, \( I \) is the current through the resistor in amperes, and \( R \) is the resistance in ohms.
This formula shows that the power dissipated is proportional to the square of the current, which means that even small changes in current can lead to significant changes in the thermal energy generated.
In the exercise, this concept is vital for understanding how much power would be lost in the human body when subjected to a dangerous current. The calculated power, 10.1834W, while not inherently large, can be significant when considering the biological impacts on the human body, indicating potential harm when exposed to such conditions.
Cylinder cross-sectional area
The cross-sectional area of a cylinder is essential for various calculations related to the flow of electricity through it. It is particularly important in determining resistance when combined with the length of the cylinder and the material's resistivity. The formula for the cross-sectional area \( A \) of a cylinder is:
  • \( A = \pi \left(\frac{d}{2}\right)^2 \)
Where \( d \) is the diameter of the cylinder.
In our exercise, calculating this area was necessary to find the body's resistance as a cylindrical conductor between two points.
By understanding the geometry and applying this formula, we were able to calculate the cross-sectional area accurately, allowing for further calculations involving bulk resistivity and resulting resistance.
This concept is equally applicable in various engineering fields and physics problems, making it a foundational part of understanding material properties in different shapes.

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Most popular questions from this chapter

When a resistor with resistance \(R\) is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that \(R\) remains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes 5.00 W. What is the voltage of this battery?

The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce \(3.15 \times 10^4\) J of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

A strand of wire has resistance 5.60 \(\mu \Omega\). Find the net resistance of 120 such strands if they are (a) placed side by side to form a cable of the same length as a single strand, and (b) connected end to end to form a wire 120 times as long as a single strand.

The current in a wire varies with time according to the relationship \(I = 55 A\) - \(10.65 A/s{^2}2t{^2}\). (a) How many coulombs of charge pass a cross section of the wire in the time interval between \(t =\) 0 and \(t =\) 8.0 s? (b) What constant current would transport the same charge in the same time interval?

In an ionic solution, a current consists of Ca\(^2+\) ions (of charge \(+2e\)) and Cl\(^-\) ions (of charge \(-e\)) traveling in opposite directions. If \(5.11 \times 1018\) Cl\(^-\) ions go from \(A\) to \(B\) every 0.50 min, while 3.24 \(\times\) 10\(^{18}\) Ca\(^2+\) ions move from \(B\) to \(A\), what is the current (in mA) through this solution, and in which direction (from \(A\) to \(B\) or from \(B\) to \(A\)) is it going?

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