Two wires of resistance `R_(1)` and `R_(2)` have temperature coefficient of resistance `alpha_(1)` and `alpha_(2)` respectively. These are joined in series. The effective temperature coefficient of resistance is

Options

A. (alpha_(1) + alpha_(2))/2`
B. `(sqrt(alpha_(1)alpha_2))`
C. `(alpha_(1)R_(1) + alpha_(2)R_(2))/(R_(1) + R_(2))`
D. `(sqrt(R_(1)R_(2)alpha_(1)alpha_(2))/(sqrt(R_(1)^(2) + R_(2)^(2))`

Answer

To find the effective temperature coefficient of resistance for two wires joined in series, we can follow these steps:

### Step 1: Understand the relationship between resistance and temperature
The resistance \( R \) of a material changes with temperature according to the formula:
\[ R(T) = R_0(1 + \alpha (T - T_0)) \]
where \( R_0 \) is the resistance at a reference temperature \( T_0 \), \( \alpha \) is the temperature coefficient of resistance, and \( T \) is the temperature.

### Step 2: Define the resistances and their temperature coefficients
Let the resistances of the two wires be \( R_1 \) and \( R_2 \), and their respective temperature coefficients be \( \alpha_1 \) and \( \alpha_2 \).

### Step 3: Find the equivalent resistance when wires are in series
When the two resistances are connected in series, the total or equivalent resistance \( R_{eq} \) is given by:
\[ R_{eq} = R_1 + R_2 \]

### Step 4: Differentiate the equivalent resistance with respect to temperature
To find the effective temperature coefficient \( \alpha_{eq} \), we need to differentiate the equivalent resistance with respect to temperature:
\[ \frac{dR_{eq}}{dT} = \frac{dR_1}{dT} + \frac{dR_2}{dT} \]

### Step 5: Substitute the relationship for \( \frac{dR}{dT} \)
From the relationship defined earlier, we know:
\[ \frac{dR_1}{dT} = \alpha_1 R_1 \quad \text{and} \quad \frac{dR_2}{dT} = \alpha_2 R_2 \]
Thus, we can write:
\[ \frac{dR_{eq}}{dT} = \alpha_1 R_1 + \alpha_2 R_2 \]

### Step 6: Relate the effective temperature coefficient to the equivalent resistance
The effective temperature coefficient \( \alpha_{eq} \) is defined as:
\[ \alpha_{eq} = \frac{1}{R_{eq}} \frac{dR_{eq}}{dT} \]
Substituting the expression from Step 5:
\[ \alpha_{eq} = \frac{\alpha_1 R_1 + \alpha_2 R_2}{R_{eq}} \]

### Step 7: Substitute \( R_{eq} \) into the equation
Since \( R_{eq} = R_1 + R_2 \), we can substitute this into our equation:
\[ \alpha_{eq} = \frac{\alpha_1 R_1 + \alpha_2 R_2}{R_1 + R_2} \]

### Final Answer
The effective temperature coefficient of resistance for the two wires joined in series is:
\[ \alpha_{eq} = \frac{\alpha_1 R_1 + \alpha_2 R_2}{R_1 + R_2} \]

---