Problem 71 - University Physics with Modern Physics

Question

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 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 \(\mathrm{mA}\) ? (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?

Step-by-Step Solution

Answer

(a) Resistance is about 1019 Ω; (b) Potential difference for lethal shock is about 102 V; (c) Power is approximately 10.2 W.

Step 1: Calculate the Cross-Sectional Area of the Cylinder

The cylinder representing the conducting path between hands has a diameter of 0.10 m. We first need to find its radius: \[ r = \frac{0.10}{2} = 0.05 \mathrm{m} \]The cross-sectional area \( A \) of the cylinder is given by the formula for the area of a circle: \[ A = \pi r^2 = \pi (0.05)^2 = 0.00785 \mathrm{m}^2 \]

Step 2: Calculate the Resistance of the Cylinder

Using the formula for resistance \( R \) for a cylindrical object, we have: \[ R = \frac{\rho L}{A} \]Where \( \rho = 5.0 \Omega \cdot \mathrm{m} \) is the resistivity, \( L = 1.6 \mathrm{m} \) is the length of the cylinder, and \( A = 0.00785 \mathrm{m}^2 \) is the cross-sectional area. Substituting the given values, we get: \[ R = \frac{5.0 \times 1.6}{0.00785} \approx 1019.108 \Omega \] Thus, the resistance between the hands is approximately \( 1019 \Omega \).

Step 3: Calculate the Potential Difference for a Lethal Shock Current

For a 100 mA current (\( I = 0.1 \mathrm{A} \)), use Ohm's Law \( V = IR \) to find the potential difference \( V \):\[ V = 0.1 \times 1019.108 \approx 101.91 \mathrm{V} \]Thus, a potential difference of approximately 102 V results in a lethal current.

Step 4: Calculate the Power Dissipated in the Body

The power \( P \) dissipated in the body can be calculated using the formula \( P = I^2 R \). Substituting the values, we have:\[ P = (0.1)^2 \times 1019.108 \approx 10.19 \mathrm{W} \]Thus, the power dissipated in the body is approximately \( 10.2 \mathrm{W} \).

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in electrical engineering and physics that describes the relationship between the voltage, current, and resistance in an electrical circuit. It is expressed with the simple formula: \( V = IR \), where:

• \( V \) is the voltage (in volts),
• \( I \) is the current (in amperes), and
• \( R \) is the resistance (in ohms).

This law tells us that the voltage across a device is equal to the product of the current flowing through the device and its resistance.
This is critical in understanding how much voltage is needed to drive a certain amount of current through a resistor. In the exercise example, we applied Ohm's Law to calculate the potential difference needed to create a lethal current of 100 mA through the body. Using the resistance between the hands, calculated as approximately 1019 ohms, we determined that about 102 volts are necessary to drive this dangerous current.
Understanding Ohm's Law is vital for safe circuit design, ensuring the current and voltage are appropriate for the devices involved.

cross-sectional area

The cross-sectional area is a key parameter in determining the resistance of an object, especially for conducting paths modeled as cylinders. It is the area of the section cut perpendicular to the length of the material. For a cylindrical object, like the human body's conducting pathway, the cross-sectional area \( A \) can be computed using:

• \( A = \pi r^2 \)

where \( r \) is the radius of the cylinder, which is half of the diameter.
In the given problem, the diameter was 0.10 meters, leading to a radius of 0.05 meters. Substituting in this radius, the cross-sectional area was calculated as approximately \( 0.00785 \ \text{m}^2 \).
Determining the cross-sectional area is essential because it affects how easily current can pass through the conductor. A smaller cross-sectional area means less space for current to travel, hence a higher resistance. It plays a crucial role in Step 1 of the solution, forming the basis to calculate the resistance in Step 2.

cylindrical resistance

Resistance of a cylindrical object depends on its resistivity, length, and cross-sectional area. The formula to compute the resistance \( R \) for a cylindrical object is:

• \( R = \frac{\rho L}{A} \)

where \( \rho \) is the resistivity of the material, \( L \) is the length, and \( A \) is the cross-sectional area.
In the exercise scenario, the resistivity of the human body was given as \( 5.0 \ \Omega \cdot \text{m} \), and with a length of 1.6 meters and an area of \( 0.00785 \ \text{m}^2 \), we calculated the cylindrical resistance to be approximately 1019 ohms.
The cylindrical shape models the pathway of current between hands soaking in salt water, essentially making skin resistance negligible. Correctly calculating cylindrical resistance allows us to proceed with Ohm's Law calculations, providing insight into current flow and voltage requirements for safety analysis.

power dissipation

Power dissipation in an electrical circuit is the process by which electric energy is converted into heat energy. It occurs whenever current flows through a resistance and can be calculated using the formula:

• \( P = I^2 R \)

where \( P \) is the power (in watts), \( I \) is the current (in amperes), and \( R \) is the resistance (in ohms).
For the given problem, the current was 0.1 amperes and with the calculated resistance of about 1019 ohms, we found the power dissipated to be approximately 10.2 watts.
This is significant because in electronic components, high power dissipation can cause overheating, potentially damaging the component or leading to safety hazards. In biological systems, like the human body, excessive power dissipation can cause injuries or be fatal, highlighting the importance of managing current and resistance effectively.