- JEE Mains 2024
- Hard
- MCQ
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 _______.
Answer & Solution
(B) \(\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}\)
Step-by-step Solution
Detailed explanation
\(R_{e q}=R_1+R_2\)
\(2 R\left(1+\alpha_{e q} \Delta \theta\right)=R\left(1+\alpha_1 \Delta \theta\right)+R\left(1+\alpha_2 \Delta \theta\right)\)
\(2 R\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=2 R+\left(\alpha_1+\alpha_2\right) R \Delta \theta\)
\(\alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2}\)
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
\(2+2\left(\alpha_1+\alpha_2\right) \Delta \theta=2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta\)
\(\left(\alpha_1+\alpha_2\right) \Delta \theta=2 \alpha_{\mathrm{eq}} \Delta \theta\)
\(\alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2}\)
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