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

Since both conductors have the same resistance at $0^\circ\mathrm{C}$, let that common resistance be $R$.

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

2. Temperature coefficient of series combination

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

Now compare with the standard form: $R_s=R_{s0}(1+\alpha_s t)$ where $R_{s0}=2R$

Thus, $2R(1+\alpha_s t)=2R+R(\alpha_1+\alpha_2)t$ Dividing by $2R$, $1+\alpha_s t=1+\frac{\alpha_1+\alpha_2}{2}t$ Hence, $\boxed{\alpha_s=\frac{\alpha_1+\alpha_2}{2}}$

3. Temperature coefficient of parallel combination

For parallel combination, $R_p=\frac{R_1R_2}{R_1+R_2}$ Substitute: $R_p=\frac{R^2(1+\alpha_1 t)(1+\alpha_2 t)}{R[(1+\alpha_1 t)+(1+\alpha_2 t)]}$ $R_p=R\cdot \frac{(1+\alpha_1 t)(1+\alpha_2 t)}{2+(\alpha_1+\alpha_2)t}$

Since temperature coefficient is defined for small temperature variation, neglect the product term $\alpha_1\alpha_2 t^2$: $(1+\alpha_1 t)(1+\alpha_2 t)\approx 1+(\alpha_1+\alpha_2)t$ So, $R_p\approx R\cdot \frac{1+(\alpha_1+\alpha_2)t}{2+(\alpha_1+\alpha_2)t}$

Now factor 2 from denominator: $R_p\approx \frac{R}{2}\cdot \frac{1+(\alpha_1+\alpha_2)t}{1+\frac{\alpha_1+\alpha_2}{2}t}$

Using $\dfrac{1+xt}{1+yt}\approx 1+(x-y)t$ for small $t$, $R_p\approx \frac{R}{2}\left[1+\left((\alpha_1+\alpha_2)-\frac{\alpha_1+\alpha_2}{2}\right)t\right]$ $R_p\approx \frac{R}{2}\left[1+\frac{\alpha_1+\alpha_2}{2}t\right]$

Comparing with $R_p=R_{p0}(1+\alpha_p t), \qquad R_{p0}=\frac{R}{2}$ we get $\boxed{\alpha_p=\frac{\alpha_1+\alpha_2}{2}}$

4. Final result

• Series combination coefficient: $\boxed{\frac{\alpha_1+\alpha_2}{2}}$
• Parallel combination coefficient: $\boxed{\frac{\alpha_1+\alpha_2}{2}}$

Therefore the correct option is: $\boxed{\text{B}}$