Question 159: Current Electricity

Question

Two conductors have the same resistances at $$0^{\circ} \mathrm{C}$$ but their temperature coefficients of resistance are $$\alpha_1$$ and $$\alpha_2$$. The respective temperature coefficients for their series and parallel combinations are :

Accepted Answer

Series :
$$\begin{aligned}
& \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \
& 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta 	hetaight)=\mathrm{R}\left(1+\alpha_1 \Delta 	hetaight)+\mathrm{R}\left(1+\alpha_2 \Delta 	hetaight) \
& 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta 	hetaight)=2 \mathrm{R}+\left(\alpha_1+\alpha_2ight) \mathrm{R} \Delta 	heta \
& \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2}
\end{aligned}$$
Parallel :
$$\begin{aligned}
& \frac{1}{R_{	ext {eq }}}=\frac{1}{R_1}+\frac{1}{R_2} \
& \frac{1}{\frac{R}{2}\left(1+\alpha_{	ext {eq }} \Delta 	hetaight)}=\frac{1}{R\left(1+\alpha_1 \Delta 	hetaight)}+\frac{1}{\mathrm{R}\left(1+\alpha_2 \Delta 	hetaight)}
\end{aligned}$$
$$\begin{aligned}
& \frac{2}{1+\alpha_{\mathrm{eq}} \Delta 	heta}=\frac{1}{1+\alpha_1 \Delta 	heta}+\frac{1}{1+\alpha_2 \Delta 	heta} \
& \frac{2}{1+\alpha_{\mathrm{eq}} \Delta 	heta}=\frac{1+\alpha_2 \Delta 	heta+1+\alpha_1 \Delta 	heta}{\left(1+\alpha_1 \Delta 	hetaight)\left(1+\alpha_2 \Delta 	hetaight)} \
& 2\left[\left(1+\alpha_1 \Delta 	hetaight)\left(1+\alpha_2 \Delta 	hetaight)ight] \
& =\left[2+\left(\alpha_1+\alpha_2ight) \Delta 	hetaight]\left[1+\alpha_{\mathrm{eq}} \Delta 	hetaight] \
& 2\left[1+\alpha_1 \Delta 	heta+\alpha_2 \Delta 	heta+\alpha_1 \alpha_2 \Delta 	hetaight]
\end{aligned}$$
$$\begin{aligned}
& = \
& 2+2 \alpha_{\mathrm{eq}} \Delta 	heta+\left(\alpha_1+\alpha_2ight) \Delta 	heta+\alpha_{\mathrm{eq}}\left(\alpha_1+\alpha_2ight) \Delta 	heta^2
\end{aligned}$$
Neglecting small terms
$$\begin{aligned}
& 2+2\left(\alpha_1+\alpha_2ight) \Delta 	heta=2+2 \alpha_{\mathrm{eq}} \Delta 	heta+\left(\alpha_1+\alpha_2ight) \Delta 	heta \
& \left(\alpha_1+\alpha_2ight) \Delta 	heta=2 \alpha_{\mathrm{eq}} \Delta 	heta \
& \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2}
\end{aligned}$$

Answer

$$\alpha_1+\alpha_2, \frac{\alpha_1 \alpha_2}{\alpha_1+\alpha_2}$$

Answer

$$\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}$$

Answer

$$\alpha_1+\alpha_2, \frac{\alpha_1+\alpha_2}{2}$$

Answer

$$\frac{\alpha_1+\alpha_2}{2}, \alpha_1+\alpha_2$$

Current Electricity

Two conductors have the same resistances at $0^{\circ} \mathrm{C} but their temperature coefficients of resistance are \alpha_1 and \alpha_2$. The respective temperature coefficients for their series and parallel combinations are :

Choose the correct answer: