A `4 Omega` wire is doubled on it. Calculate the new resistance of the wire.

Explanation

To solve the problem of finding the new resistance of a wire when it is doubled upon itself, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Resistance**:
- We are given that the resistance of the original wire (R) is 4 ohms.

2. **Identify the Length and Cross-Sectional Area**:
- Let the length of the original wire be \( L \) and the cross-sectional area be \( A \).

3. **Use the Resistance Formula**:
- The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
- Where \( \rho \) is the resistivity of the material.

4. **Determine the New Length and Area**:
- When the wire is doubled upon itself, the new length \( L' \) becomes:
\[ L' = \frac{L}{2} \]
- The new cross-sectional area \( A' \) becomes:
\[ A' = 2A \]

5. **Calculate the New Resistance**:
- Using the resistance formula for the new configuration:
\[ R' = \frac{\rho L'}{A'} \]
- Substitute \( L' \) and \( A' \):
\[ R' = \frac{\rho \left(\frac{L}{2}\right)}{2A} \]
- Simplifying this gives:
\[ R' = \frac{\rho L}{4A} \]

6. **Substitute the Original Resistance**:
- From the original resistance formula, we know:
\[ R = \frac{\rho L}{A} = 4 \, \text{ohms} \]
- Therefore, we can express \( \frac{\rho L}{A} \) as 4:
\[ R' = \frac{4}{4} = 1 \, \text{ohm} \]

7. **Final Answer**:
- The new resistance of the wire after it has been doubled upon itself is:
\[ R' = 1 \, \text{ohm} \]