Two conductors have the same resistance at `0^@C` but their temperature coefficient of resistanc are `alpha_1 and alpha_2`. The respective temperature coefficients of their series and parallel combinations are nearly

Options

A. `(alpha_1+alpha_2)/2, alpha_1 + alpha_2`
B. `alpha_1+alpha_2, (alpha_1+alpha_2)/2`
C. `alpha_1+alpha_2, (alpha_1alpha_2)/(alpha_1+alpha_2)`
D. `(alpha_1+alpha_2)/2, (alpha_1+alpha_2)/2`

Explanation

To solve the problem, we need to find the effective temperature coefficients of resistance for two conductors when they are connected in series and in parallel. Let's denote the resistances of the two conductors at \(0^\circ C\) as \(R_0\) and their temperature coefficients of resistance as \(\alpha_1\) and \(\alpha_2\).

### Step 1: Find the resistance of the conductors at temperature \(T\)

For a conductor, the resistance at a temperature \(T\) is given by:
\[ R(T) = R_0(1 + \alpha T) \]
Thus, for the two conductors, the resistances at temperature \(T\) will be:
- For conductor 1:
\[ R_1(T) = R_0(1 + \alpha_1 T) \]
- For conductor 2:
\[ R_2(T) = R_0(1 + \alpha_2 T) \]

### Step 2: Calculate the total resistance in series

When the two conductors are connected in series, the total resistance \(R_s\) is:
\[ R_s = R_1(T) + R_2(T) = R_0(1 + \alpha_1 T) + R_0(1 + \alpha_2 T) \]
\[ R_s = R_0(2 + \alpha_1 T + \alpha_2 T) \]

### Step 3: Find the effective temperature coefficient for series combination

The effective temperature coefficient \(\alpha_s\) for the series combination can be derived from the total resistance:
\[ R_s = R_0(2 + \alpha_s T) \]
Equating the two expressions for \(R_s\):
\[ R_0(2 + \alpha_s T) = R_0(2 + (\alpha_1 + \alpha_2) T) \]
Dividing both sides by \(R_0\) and rearranging gives:
\[ \alpha_s = \frac{\alpha_1 + \alpha_2}{2} \]

### Step 4: Calculate the total resistance in parallel

When the two conductors are connected in parallel, the total resistance \(R_p\) is given by:
\[ \frac{1}{R_p} = \frac{1}{R_1(T)} + \frac{1}{R_2(T)} \]
Substituting the expressions for \(R_1(T)\) and \(R_2(T)\):
\[ \frac{1}{R_p} = \frac{1}{R_0(1 + \alpha_1 T)} + \frac{1}{R_0(1 + \alpha_2 T)} \]
Combining the fractions:
\[ \frac{1}{R_p} = \frac{(1 + \alpha_2 T) + (1 + \alpha_1 T)}{R_0(1 + \alpha_1 T)(1 + \alpha_2 T)} \]
\[ \frac{1}{R_p} = \frac{2 + (\alpha_1 + \alpha_2) T}{R_0(1 + \alpha_1 T)(1 + \alpha_2 T)} \]

### Step 5: Find the effective temperature coefficient for parallel combination

To find the effective temperature coefficient \(\alpha_p\), we can express \(R_p\) in a similar form:
\[ R_p = \frac{R_0(1 + \alpha_1 T)(1 + \alpha_2 T)}{2 + (\alpha_1 + \alpha_2) T} \]
Using the approximation for small values of \(\alpha_1\) and \(\alpha_2\), we can simplify this expression to find \(\alpha_p\):
\[ \alpha_p \approx \frac{\alpha_1 + \alpha_2}{2} \]

### Final Result

Thus, the effective temperature coefficients of resistance for the series and parallel combinations are:
- For series:
\[ \alpha_s = \frac{\alpha_1 + \alpha_2}{2} \]
- For parallel:
\[ \alpha_p = \frac{\alpha_1 + \alpha_2}{2} \]