JEE Main 2024 — Physics Thermodynamics

Change in resistance due to temperature is given by, ${R}^{'}=R(1+\alpha \Delta T)$R′=R(1+αΔT)

For Series Combination:

$${R}_{eq}={R}_{1}+{R}_{2}$$Req​=R1​+R2​

$$\Rightarrow 2R(1+{\alpha }_{eq}\Delta T)=R(1+{\alpha }_{1}\Delta T)+R(1+{\alpha }_{2}\Delta T)$$⇒2R(1+αeq​ΔT)=R(1+α1​ΔT)+R(1+α2​ΔT)

$$\Rightarrow 2R(1+{\alpha }_{eq}\Delta T)=2R+({\alpha }_{1}+{\alpha }_{2})R\Delta T$$⇒2R(1+αeq​ΔT)=2R+(α1​+α2​)RΔT

$$\Rightarrow {\alpha }_{eq}=\frac{{\alpha }_{1}+{\alpha }_{2}}{2}$$⇒αeq​=2α1​+α2​​

For Parallel Combination:

$$\frac{1}{{R}_{eq}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}$$Req​1​=R1​1​+R2​1​

$$\Rightarrow \frac{1}{\frac{R}{2}(1+{\alpha }_{eq}\Delta T)}=\frac{1}{R(1+{\alpha }_{1}\Delta T)}+\frac{1}{R(1+{\alpha }_{2}\Delta T)}$$⇒2R​(1+αeq​ΔT)1​=R(1+α1​ΔT)1​+R(1+α2​ΔT)1​

$$\Rightarrow \frac{2}{1+{\alpha }_{eq}\Delta T}=\frac{1}{1+{\alpha }_{1}\Delta T}+\frac{1}{1+{\alpha }_{2}\Delta T}$$⇒1+αeq​ΔT2​=1+α1​ΔT1​+1+α2​ΔT1​

$$\Rightarrow \frac{2}{1+{\alpha }_{eq}\Delta T}=\frac{1+{\alpha }_{2}\Delta T+1+{\alpha }_{1}\Delta T}{(1+{\alpha }_{1}\Delta T)(1+{\alpha }_{2}\Delta T)}$$⇒1+αeq​ΔT2​=(1+α1​ΔT)(1+α2​ΔT)1+α2​ΔT+1+α1​ΔT​

$$\Rightarrow 2[(1+{\alpha }_{1}\Delta T)(1+{\alpha }_{2}\Delta T)]=[2+({\alpha }_{1}+{\alpha }_{2})\Delta T][1+{\alpha }_{eq}\Delta T]$$⇒2[(1+α1​ΔT)(1+α2​ΔT)]=[2+(α1​+α2​)ΔT][1+αeq​ΔT]

$$\Rightarrow 2[1+{\alpha }_{1}\Delta T+{\alpha }_{2}\Delta T+{\alpha }_{1}{\alpha }_{2}\Delta T]=2+2{\alpha }_{eq}\Delta T+({\alpha }_{1}+{\alpha }_{2})\Delta T+{\alpha }_{eq}({\alpha }_{1}+{\alpha }_{2})\Delta {T}^{2}$$⇒2[1+α1​ΔT+α2​ΔT+α1​α2​ΔT]=2+2αeq​ΔT+(α1​+α2​)ΔT+αeq​(α1​+α2​)ΔT2

Neglecting small terms

$$2+2({\alpha }_{1}+{\alpha }_{2})\Delta T=2+2{\alpha }_{eq}\Delta T+({\alpha }_{1}+{\alpha }_{2})\Delta T$$2+2(α1​+α2​)ΔT=2+2αeq​ΔT+(α1​+α2​)ΔT

$$\Rightarrow ({\alpha }_{1}+{\alpha }_{2})\Delta T=2{\alpha }_{eq}\Delta T$$⇒(α1​+α2​)ΔT=2αeq​ΔT

$$\Rightarrow {\alpha }_{eq}=\frac{{\alpha }_{1}+{\alpha }_{2}}{2}$$⇒αeq​=2α1​+α2​​