An aluminium `(alpha_(Al)=4xx 10^(-3)//^(0)C)` wire resis-tance `.R_(1).` and carbon wire `(alpha_(c)=0.5 xx 10^(-3)//^(0)C)` resistance `.R_(2).` are connected in series to have a resultant resistance of 18 ohm at all temperatures The values of `R_(1)` and `R_(2)` in ohms

Options

A. 2, 16
B. 12, 6
C. 13, 5
D. 14, 4

Answer

To solve the problem, we need to find the resistances \( R_1 \) (for the aluminum wire) and \( R_2 \) (for the carbon wire) given that they are connected in series and the total resistance is 18 ohms.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two wires: one made of aluminum and the other made of carbon.
- The temperature coefficients of resistance are given as:
- \( \alpha_{Al} = 4 \times 10^{-3} \, \text{°C}^{-1} \) for aluminum.
- \( \alpha_{C} = 0.5 \times 10^{-3} \, \text{°C}^{-1} \) for carbon.
- The total resistance \( R_{total} = R_1 + R_2 = 18 \, \text{ohms} \).

2. **Setting Up the Equation**:
- The change in resistance due to temperature for each wire can be expressed as:
\[ R_1' = R_1(1 + \alpha_{Al} \Delta T) \]
\[ R_2' = R_2(1 - \alpha_{C} \Delta T) \]
- Since the total resistance remains constant at all temperatures, we can write:
\[ R_1' + R_2' = 18 \]

3. **Substituting the Resistance Changes**:
- Substituting the expressions for \( R_1' \) and \( R_2' \):
\[ R_1(1 + \alpha_{Al} \Delta T) + R_2(1 - \alpha_{C} \Delta T) = 18 \]

4. **Simplifying the Equation**:
- Rearranging gives:
\[ R_1 + R_2 + R_1 \alpha_{Al} \Delta T - R_2 \alpha_{C} \Delta T = 18 \]
- We know \( R_1 + R_2 = 18 \), so we can cancel this term:
\[ R_1 \alpha_{Al} \Delta T - R_2 \alpha_{C} \Delta T = 0 \]

5. **Factoring Out \( \Delta T \)**:
- Factoring out \( \Delta T \) gives:
\[ \Delta T (R_1 \alpha_{Al} - R_2 \alpha_{C}) = 0 \]
- Since \( \Delta T \) cannot be zero, we have:
\[ R_1 \alpha_{Al} = R_2 \alpha_{C} \]

6. **Expressing \( R_2 \) in Terms of \( R_1 \)**:
- Rearranging gives:
\[ R_2 = \frac{R_1 \alpha_{Al}}{\alpha_{C}} \]

7. **Substituting Values**:
- Substituting the values of \( \alpha_{Al} \) and \( \alpha_{C} \):
\[ R_2 = \frac{R_1 \cdot 4 \times 10^{-3}}{0.5 \times 10^{-3}} = 8 R_1 \]

8. **Using the Total Resistance**:
- Now substituting \( R_2 \) back into the total resistance equation:
\[ R_1 + 8 R_1 = 18 \]
\[ 9 R_1 = 18 \]
\[ R_1 = 2 \, \text{ohms} \]

9. **Finding \( R_2 \)**:
- Now substituting \( R_1 \) back to find \( R_2 \):
\[ R_2 = 8 R_1 = 8 \times 2 = 16 \, \text{ohms} \]

### Final Answers:
- \( R_1 = 2 \, \text{ohms} \)
- \( R_2 = 16 \, \text{ohms} \)