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

(B) $\frac{\alpha_{1}+\alpha_{2}}{2}, \frac{\alpha_{1}+\alpha_{2}}{2}$

(A) $\alpha_{1}+\alpha_{2}, \frac{\alpha_{1}+\alpha_{2}}{2}$

(C) $\alpha_{1}+\alpha_{2}, \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}}$

(D) $\frac{\alpha_{1}+\alpha_{2}}{2}, \alpha_{1}+\alpha_{2}$

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Current Electricity

Two conductors have the same resistances at

0^{\circ} C

but their temperature coefficients of resistance are

α_{1}

and

α_{2}

. The respective temperature coefficients for their series and parallel combinations are:

Answer: (B)

Let both conductors have resistance

R

0^{\circ}

C.

Series:

R_{e q} = R_{1} + R_{2} = R (1 + α_{1} Δ θ) + R (1 + α_{2} Δ θ)

2 R (1 + α_{e q} Δ θ) = 2 R + R (α_{1} + α_{2}) Δ θ ⟹ α_{e q} = \frac{α _{1} + α _{2}}{2}

Parallel:

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

. Expanding to first order in

Δ θ

gives the same result:

α_{e q} = \frac{α _{1} + α _{2}}{2}

So both the series and parallel combinations have temperature coefficient

\frac{α _{1} + α _{2}}{2}

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option