The resistivity of copper at room temperature is 1.7x10-8 Qm. A copper wire of 100 m long and 2 mm in diameter is connected to a power source with a potential difference of 15 V. Find the current. I =
The resistivity of copper at room temperature is 1.7x10-8 Qm. A copper wire of 100 m long and 2 mm in diameter is connected to a power source with a potential difference of 15 V. Find the current. I =
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![**Problem Statement:**
The resistivity of copper at room temperature is \(1.7 \times 10^{-8} \, \Omega \cdot \text{m}\). A copper wire of 100 m long and 2 mm in diameter is connected to a power source with a potential difference of 15 V. Find the current.
\[ I = \_\_\_ \text{A} \]
**Explanation:**
To find the current, \(I\), flowing through the wire, you can use Ohm’s Law and the formula for resistance in terms of resistivity.
1. **Calculate the cross-sectional area of the wire:**
The diameter of the wire is 2 mm, which is equal to 0.002 m. The radius is half of the diameter, so:
\[
\text{Radius} = \frac{0.002}{2} = 0.001 \, \text{m}
\]
The cross-sectional area, \(A\), of a circle is given by:
\[
A = \pi r^2 = \pi (0.001)^2 = \pi \times 10^{-6} \, \text{m}^2
\]
2. **Calculate the resistance, \(R\):**
The resistance of a wire is given by:
\[
R = \frac{\rho L}{A}
\]
Where:
- \(\rho = 1.7 \times 10^{-8} \, \Omega \cdot \text{m}\) (resistivity)
- \(L = 100 \, \text{m}\) (length of the wire)
- \(A = \pi \times 10^{-6} \, \text{m}^2\) (cross-sectional area)
Substitute these values into the formula:
\[
R = \frac{1.7 \times 10^{-8} \times 100}{\pi \times 10^{-6}} \, \Omega
\]
3. **Apply Ohm’s Law to find the current, \(I\):**
Ohm’s Law states:
\[
V = IR
\]
Rearrange to solve for \(I\):
\[
I = \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F73a49f30-5066-4bcb-bfde-0b28dc9bcf6a%2F603dc777-1457-4841-9c2e-5c86717a4726%2Fbivxaif_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
The resistivity of copper at room temperature is \(1.7 \times 10^{-8} \, \Omega \cdot \text{m}\). A copper wire of 100 m long and 2 mm in diameter is connected to a power source with a potential difference of 15 V. Find the current.
\[ I = \_\_\_ \text{A} \]
**Explanation:**
To find the current, \(I\), flowing through the wire, you can use Ohm’s Law and the formula for resistance in terms of resistivity.
1. **Calculate the cross-sectional area of the wire:**
The diameter of the wire is 2 mm, which is equal to 0.002 m. The radius is half of the diameter, so:
\[
\text{Radius} = \frac{0.002}{2} = 0.001 \, \text{m}
\]
The cross-sectional area, \(A\), of a circle is given by:
\[
A = \pi r^2 = \pi (0.001)^2 = \pi \times 10^{-6} \, \text{m}^2
\]
2. **Calculate the resistance, \(R\):**
The resistance of a wire is given by:
\[
R = \frac{\rho L}{A}
\]
Where:
- \(\rho = 1.7 \times 10^{-8} \, \Omega \cdot \text{m}\) (resistivity)
- \(L = 100 \, \text{m}\) (length of the wire)
- \(A = \pi \times 10^{-6} \, \text{m}^2\) (cross-sectional area)
Substitute these values into the formula:
\[
R = \frac{1.7 \times 10^{-8} \times 100}{\pi \times 10^{-6}} \, \Omega
\]
3. **Apply Ohm’s Law to find the current, \(I\):**
Ohm’s Law states:
\[
V = IR
\]
Rearrange to solve for \(I\):
\[
I = \
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