\(0^{\circ} \mathrm{C}\) તાપમાને બે સુવાહકોના અવરોધ સમાન છે. આ સુવાહકો માટે અવરોધકતા તાપમાન ગુણાંક \(\alpha_1\) અને \(\alpha_2\) છે. તો આપેલ સુવાહકોના શ્રેણી અને સમાંતર જોડાણ માટે સમતુલ્ય અવરોધકતા તાપમાન ગુણાંક _______ છે.

\(R_{e q}=R_1+R_2\)

\(2 R\left(1+\alpha_{e q} \Delta \theta\right)=R\left(1+\alpha_1 \Delta \theta\right)+R\left(1+\alpha_2 \Delta \theta\right)\)

\(2 R\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=2 R+\left(\alpha_1+\alpha_2\right) R \Delta \theta\)

\(\alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2}\)

Parallel :

\(\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}\)

\(\frac{1}{\frac{R}{2}\left(1+\alpha_{e q} \Delta \theta\right)}=\frac{1}{R\left(1+\alpha_1 \Delta \theta\right)}+\frac{1}{R\left(1+\alpha_2 \Delta \theta\right)}\)

\(\frac{2}{1+\alpha_{\mathrm{eq}} \Delta \theta}=\frac{1}{1+\alpha_1 \Delta \theta}+\frac{1}{1+\alpha_2 \Delta \theta}\)

\(\frac{2}{1+\alpha_{\mathrm{eq}} \Delta \theta}=\frac{1+\alpha_2 \Delta \theta+1+\alpha_1 \Delta \theta}{\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)}\)

\(2\left[\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)\right]\)

\(=\left[2+\left(\alpha_1+\alpha_{22}\right) \Delta \theta\right]\left[1+\alpha_{\mathrm{eq}} \Delta \theta\right]\)

\(2\left[1+\alpha_1 \Delta \theta+\alpha_2 \Delta \theta+\alpha_1 \alpha_2 \Delta \theta\right]\)

\(=2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_{22}\right) \Delta \theta+\alpha_{\mathrm{eq}}\left(\alpha_1+\alpha_2\right) \Delta \theta^2\)

Neglecting small terms

\(2+2\left(\alpha_1+\alpha_2\right) \Delta \theta=2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta\)

\(\left(\alpha_1+\alpha_2\right) \Delta \theta=2 \alpha_{\mathrm{eq}} \Delta \theta\)

\(\alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2}\)