1. Write resistance of each conductor at temperature $t$

Since both have the same resistance at $0^\circ C$, let that resistance be $R$.

Then at temperature $t$, $R_1 = R(1+\alpha_1 t), \qquad R_2 = R(1+\alpha_2 t)$

For temperature coefficient of a combination, we compare the total resistance with its value at $0^\circ C$.

2. Series combination

In series, $R_s = R_1 + R_2 = R(1+\alpha_1 t)+R(1+\alpha_2 t)$ $R_s = 2R + R(\alpha_1+\alpha_2)t$

At $0^\circ C$, series resistance is $R_{s0}=2R$

Let the temperature coefficient of the series combination be $\alpha_s$. Then $R_s = R_{s0}(1+\alpha_s t)=2R(1+\alpha_s t)$

Comparing with $R_s = 2R + R(\alpha_1+\alpha_2)t$ we get $2R\alpha_s t = R(\alpha_1+\alpha_2)t$ $\alpha_s = \frac{\alpha_1+\alpha_2}{2}$

3. Parallel combination

In parallel, $R_p = \frac{R_1R_2}{R_1+R_2}$

Substitute: $R_p = \frac{R(1+\alpha_1 t)\,R(1+\alpha_2 t)}{R(1+\alpha_1 t)+R(1+\alpha_2 t)}$ $R_p = R\,\frac{(1+\alpha_1 t)(1+\alpha_2 t)}{2+(\alpha_1+\alpha_2)t}$

Since the question says nearly, we neglect second-order term $\alpha_1\alpha_2 t^2$: $(1+\alpha_1 t)(1+\alpha_2 t) \approx 1+(\alpha_1+\alpha_2)t$

So, $R_p \approx R\frac{1+(\alpha_1+\alpha_2)t}{2+(\alpha_1+\alpha_2)t}$

Now use $\frac{1+xt}{2+xt} = \frac{1}{2}\cdot \frac{1+xt}{1+\frac{xt}{2}}$ where $x=\alpha_1+\alpha_2$.

For small $xt$, $\frac{1}{1+\frac{xt}{2}} \approx 1-\frac{xt}{2}$

Thus, $R_p \approx \frac{R}{2}(1+xt)\left(1-\frac{xt}{2}\right)$ Neglecting higher-order terms, $R_p \approx \frac{R}{2}\left(1+\frac{xt}{2}\right)$

So, $R_p \approx \frac{R}{2}\left(1+\frac{\alpha_1+\alpha_2}{2}t\right)$

At $0^\circ C$, parallel resistance is $R_{p0} = \frac{R\cdot R}{R+R}=\frac{R}{2}$

Hence temperature coefficient of parallel combination is $\alpha_p = \frac{\alpha_1+\alpha_2}{2}$

4. Final result

Therefore, the temperature coefficients of:

• series combination $= \dfrac{\alpha_1+\alpha_2}{2}$
• parallel combination $= \dfrac{\alpha_1+\alpha_2}{2}$

So the correct option is: $\boxed{\text{D}}$

5. Comparison with stored answer

Stored correct answer: D

Derived answer: D

They agree.