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To solve the problem, we need to calculate the required cross-sectional area of a copper conductor that is 4 meters long, with a current of 144 A and a voltage of 9 V across it.

### Step 1: Using Ohm's Law

Ohm's Law relates voltage (V), current (I), and resistance (R) as:
\[
V = I \times R
\]
We can rearrange this to solve for resistance \(R\):
\[
R = \frac{V}{I} = \frac{9 \, \text{V}}{144 \, \text{A}} = 0.0625 \, \Omega
\]

### Step 2: Using the Formula for Resistance of a Wire

The resistance of a wire is also given by:
\[
R = \rho \times \frac{L}{A}
\]
Where:
- \(\rho\) is the resistivity of the material (copper in this case),
- \(L\) is the length of the wire (4 m),
- \(A\) is the cross-sectional area we need to find.

From the table, the resistivity of copper (\(\rho\)) is given as \(0.017 \, \Omega \cdot \text{mm}^2 / \text{m}\).

### Step 3: Rearranging for Cross-sectional Area

We rearrange the formula to solve for \(A\):
\[
A = \frac{\rho \times L}{R}
\]
Substituting the known values:
\[
A = \frac{0.017 \times 4}{0.0625} = \frac{0.068}{0.0625} \approx 1.088 \, \text{mm}^2
\]

Thus, the required cross-sectional area of the copper conductor is approximately **1.088 mm²**.

---

Would you like a detailed explanation of any step or have further questions?

### Relative Questions:
1. How does resistivity vary between different materials?
2. Why is copper commonly used for conductors?
3. How does the length of a conductor affect its resistance?
4. What happens to the resistance of a wire when the cross-sectional area is doubled?
5. How can we minimize power loss in electrical cables?

### Tip: 
When designing electrical systems, choosing materials with lower resistivity and proper cross-sectional areas can reduce energy losses and prevent overheating.

Мідний провідник довжиною 4 м, пропускає крізь себе силу струму 144 А, при цьому напруга дорівнює 9 В. Яка повинна бути площа поперечного перерізу провідника?

Solve the problem of calculating the cross-sectional area of a copper conductor of 4 meters, carrying a current of 144 A, with a voltage of 9 V. Learn how to apply Ohm's law and resistance formulas, using copper resistivity data. Suitable for Grades 10-12.

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