Thermal Properties of Matter

JEE Main level, single correct, medium

91 of 126

Question 91 of 126

Two conductors have the same resistances at 0∘C0^{\circ}C but their temperature coefficients of resistance are α1{\alpha }_{1} and α2{\alpha }_{2}. The respective temperature coefficients for their series and parallel combinations are :

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Answer: (B) α1+α22,α1+α22\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2},\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2}

Change in resistance due to temperature is given by, R′=R(1+α∆T){R}^{'}=R\left(1+\alpha ∆T\right)
For Series Combination:
Req=R1+R2{R}_{eq}={R}_{1}+{R}_{2}
⇒2R(1+αeq∆T)=R(1+α1∆T)+R(1+α2∆T)\Rightarrow 2R\left(1+{\alpha }_{eq}∆T\right)=R\left(1+{\alpha }_{1}∆T\right)+R\left(1+{\alpha }_{2}∆T\right)
⇒2R(1+αeq∆T)=2R+(α1+α2)R∆T\Rightarrow 2R\left(1+{\alpha }_{eq}∆T\right)=2R+\left({\alpha }_{1}+{\alpha }_{2}\right)R∆T
⇒αeq=α1+α22\Rightarrow {\alpha }_{eq}=\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2}
For Parallel Combination:
1Req=1R1+1R2\dfrac{1}{{R}_{eq}}=\dfrac{1}{{R}_{1}}+\dfrac{1}{{R}_{2}}
⇒1R2(1+αeq∆T)=1R(1+α1∆T)+1R(1+α2∆T)\Rightarrow \dfrac{1}{\dfrac{R}{2}\left(1+{\alpha }_{eq}∆T\right)}=\dfrac{1}{R\left(1+{\alpha }_{1}∆T\right)}+\dfrac{1}{R\left(1+{\alpha }_{2}∆T\right)}
⇒21+αeq∆T=11+α1∆T+11+α2∆T\Rightarrow \dfrac{2}{1+{\alpha }_{eq}∆T}=\dfrac{1}{1+{\alpha }_{1}∆T}+\dfrac{1}{1+{\alpha }_{2}∆T}
⇒21+αeq∆T=1+α2∆T+1+α1∆T(1+α1∆T)(1+α2∆T)\Rightarrow \dfrac{2}{1+{\alpha }_{eq}∆T}=\dfrac{1+{\alpha }_{2}∆T+1+{\alpha }_{1}∆T}{\left(1+{\alpha }_{1}∆T\right)\left(1+{\alpha }_{2}∆T\right)}
⇒2[(1+α1∆T)(1+α2∆T)]=[2+(α1+α2)∆T][1+αeq∆T]\Rightarrow 2\left[\left(1+{\alpha }_{1}∆T\right)\left(1+{\alpha }_{2}∆T\right)\right]=\left[2+\left({\alpha }_{1}+{\alpha }_{2}\right)∆T\right]\left[1+{\alpha }_{eq}∆T\right]
⇒2[1+α1∆T+α2∆T+α1α2∆T]=2+2αeq∆T+(α1+α2)∆T+αeq(α1+α2)∆T2\Rightarrow 2\left[1+{\alpha }_{1}∆T+{\alpha }_{2}∆T+{\alpha }_{1}{\alpha }_{2}∆T\right]=2+2{\alpha }_{eq}∆T+\left({\alpha }_{1}+{\alpha }_{2}\right)∆T+{\alpha }_{eq}\left({\alpha }_{1}+{\alpha }_{2}\right)∆{T}^{2}
Neglecting small terms
2+2(α1+α2)∆T=2+2αeq∆T+(α1+α2)∆T2+2\left({\alpha }_{1}+{\alpha }_{2}\right)∆T=2+2{\alpha }_{eq}∆T+\left({\alpha }_{1}+{\alpha }_{2}\right)∆T
⇒(α1+α2)∆T=2αeq∆T\Rightarrow \left({\alpha }_{1}+{\alpha }_{2}\right)∆T=2{\alpha }_{eq}∆T
⇒αeq=α1+α22\Rightarrow {\alpha }_{eq}=\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2}

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