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JEE MAINS 2024
JEE-Main_310124_(S1)_HT
Question
Two conductors have the same resistances at but their temperature coefficients of resistance are and . The respective temperature coefficients for their series and parallel combinations are :
Select the correct option:
A
$\alpha_1+\alpha_2, \frac{\alpha_1+\alpha_2}{2}$
B
$\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}$
C
$\alpha_1+\alpha_2, \frac{\alpha_1 \alpha_2}{\alpha_1+\alpha_2}$
D
$\frac{\alpha_1+\alpha_2}{2}, \alpha_1+\alpha_2$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol. Series : $$ \begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \\ & 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=\mathrm{R}\left(1+\alpha_1 \Delta \theta\right)+\mathrm{R}\left(1+\alpha_2 \Delta \theta\right) \\ & 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=2 \mathrm{R}+\left(\alpha_1+\alpha_2\right) \mathrm{R} \Delta \theta \end{aligned} $$ $$ \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2} $$ Parallel: $$ \begin{aligned} & \frac{1}{R_{\text {eq }}}=\frac{1}{R_1}+\frac{1}{R_2} \\ & \frac{1}{\frac{R}{2}\left(1+\alpha_{\text {eq }} \Delta \theta\right)}=\frac{1}{R\left(1+\alpha_1 \Delta \theta\right)}+\frac{1}{R\left(1+\alpha_2 \Delta \theta\right)} \\ & \frac{2}{1+\alpha_{\text {eq }} \Delta \theta}=\frac{1}{1+\alpha_1 \Delta \theta}+\frac{1}{1+\alpha_2 \Delta \theta} \\ & \frac{2}{1+\alpha_{\text {eq }} \Delta \theta}=\frac{1+\alpha_2 \Delta \theta+1+\alpha_1 \Delta \theta}{\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)} \\ & 2\left[\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)\right] \\ & =\left[2+\left(\alpha_1+\alpha_2\right) \Delta \theta\right]\left[1+\alpha_{\text {eq }} \Delta \theta\right] \\ & 2\left[1+\alpha_1 \Delta \theta+\alpha_2 \Delta \theta+\alpha_1 \alpha_2 \Delta \theta\right] \\ & =2+2 \alpha_{\text {eq }} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta+\alpha_{\text {eq }}\left(\alpha_1+\alpha_2\right) \Delta \theta^2 \end{aligned} $$ Neglecting small terms $$ \begin{aligned} & 2+2\left(\alpha_1+\alpha_2\right) \Delta \theta=2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta \\ & \left(\alpha_1+\alpha_2\right) \Delta \theta=2 \alpha_{\mathrm{eq}} \Delta \theta \\ & \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2} \end{aligned} $$
Question Tags
JEE Main
Physics
Hard
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