A wire is 0.81 m long and 0.88 mm² in cross-sectional area. It carries a current of 5.5 A when a 2.6 V potential difference is applied between its ends. Calculate the conductivity of the material of which this wire is made. Number i Units
A wire is 0.81 m long and 0.88 mm² in cross-sectional area. It carries a current of 5.5 A when a 2.6 V potential difference is applied between its ends. Calculate the conductivity of the material of which this wire is made. Number i Units
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem Statement: Electrical Conductivity Calculation**
---
**Description:**
A wire is 0.81 meters long and has a cross-sectional area of 0.88 mm². It carries a current of 5.5 amperes when a potential difference of 2.6 volts is applied between its ends. Calculate the conductivity \(\sigma\) of the material from which this wire is made.
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**Input Fields:**
- **Number:** [Input Field]
- **Units:** [Drop-down Menu]
Explanation:
1. **Length of the wire (L):** \(0.81 \, \text{m}\)
2. **Cross-sectional area (A):** \(0.88 \, \text{mm}^2\)
3. **Current (I):** \(5.5 \, \text{A}\)
4. **Potential difference (V):** \(2.6 \, \text{V}\)
To find the electrical conductivity \(\sigma\), you can use the formula for resistivity \(\rho\) and then find conductivity as \(\sigma = \frac{1}{\rho}\).
The resistance \(R\) of the wire can be calculated using Ohm's Law:
\[ R = \frac{V}{I} \]
Then, the resistivity \(\rho\) is given by:
\[ \rho = R \cdot \frac{A}{L} \]
Finally, the conductivity is:
\[ \sigma = \frac{1}{\rho} \]
Ensure that units are consistent during calculations. For instance, convert the area from mm² to m² before using it in the resistivity formula.
---
**Note:**
This problem involves understanding the relationships between electrical quantities and manipulating equations to find the desired property of the material.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2582877-f8f8-44db-83ab-009b8ab7497d%2Fc769d225-4efb-455a-9bf7-1a1c4f5f133e%2Fwp9r88d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement: Electrical Conductivity Calculation**
---
**Description:**
A wire is 0.81 meters long and has a cross-sectional area of 0.88 mm². It carries a current of 5.5 amperes when a potential difference of 2.6 volts is applied between its ends. Calculate the conductivity \(\sigma\) of the material from which this wire is made.
---
**Input Fields:**
- **Number:** [Input Field]
- **Units:** [Drop-down Menu]
Explanation:
1. **Length of the wire (L):** \(0.81 \, \text{m}\)
2. **Cross-sectional area (A):** \(0.88 \, \text{mm}^2\)
3. **Current (I):** \(5.5 \, \text{A}\)
4. **Potential difference (V):** \(2.6 \, \text{V}\)
To find the electrical conductivity \(\sigma\), you can use the formula for resistivity \(\rho\) and then find conductivity as \(\sigma = \frac{1}{\rho}\).
The resistance \(R\) of the wire can be calculated using Ohm's Law:
\[ R = \frac{V}{I} \]
Then, the resistivity \(\rho\) is given by:
\[ \rho = R \cdot \frac{A}{L} \]
Finally, the conductivity is:
\[ \sigma = \frac{1}{\rho} \]
Ensure that units are consistent during calculations. For instance, convert the area from mm² to m² before using it in the resistivity formula.
---
**Note:**
This problem involves understanding the relationships between electrical quantities and manipulating equations to find the desired property of the material.
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