A `5Omega` resistance wire is doubled on it. Calculate the new resistance of the wire.

Options

A. `1.25Omega`
B. `2.25Omega`
C. `1.5Omega`
D. `1.00Omega`

Answer

To solve the problem of finding the new resistance of a wire that has been doubled on itself, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The initial resistance of the wire is given as \( R = 5 \, \Omega \).
- Let the original length of the wire be \( L \) and the original cross-sectional area be \( A \).

2. **Determine the New Length and Area**:
- When the wire is doubled, its length becomes half of the original length. Therefore, the new length \( L' \) is:
\[ L' = \frac{L}{2} \]
- The cross-sectional area will effectively double because the wire is folded. Thus, the new area \( A' \) is:
\[ A' = 2A \]

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. **Calculate the New Resistance**:
- The new resistance \( R' \) can be calculated using the new length and area:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho \left(\frac{L}{2}\right)}{2A} \]
- Simplifying this expression:
\[ R' = \frac{\rho L}{4A} \]

5. **Relate to the Original Resistance**:
- From the original resistance \( R = \frac{\rho L}{A} = 5 \, \Omega \), we can express \( \frac{\rho L}{A} \) as 5:
\[ R' = \frac{5}{4} \, \Omega \]

6. **Final Calculation**:
- Therefore, the new resistance \( R' \) is:
\[ R' = 1.25 \, \Omega \]

### Conclusion:
The new resistance of the wire after being doubled is \( 1.25 \, \Omega \).

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