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

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

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

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

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

JEE Main · 2010 · 11 Apr 2010 · Offline · Current Electricity

Two conductors have the same resistance at

0^{\circ} C

but their temperature coefficients of resistance are

α_{1}

and

α_{2}

. The respective temperature coefficients of their series and parallel combinations are nearly

(A) $\frac{α _{1} + α _{2}}{2}, α_{1} + α_{2}$
(B) $α_{1} + α_{2}, \frac{α _{1} + α _{2}}{2}$
(C) $α_{1} + α_{2}, \frac{α _{1} α _{2}}{α _{1} + α _{2}}$
(D) $\frac{α _{1} + α _{2}}{2}, \frac{α _{1} + α _{2}}{2}$ ✓ Correct

Answer: (D)

Solution

Take both conductors to have resistance

R_{0}

0^{\circ}

C, so at temperature

θ

they read

R_{0} (1 + α_{1} θ)

and

R_{0} (1 + α_{2} θ)

. In series the total is

R_{0} (2 + (α_{1} + α_{2}) θ) = 2 R_{0} (1 + \frac{α _{1} + α _{2}}{2} θ)

, so the series combination carries the coefficient

\frac{α _{1} + α _{2}}{2}

. In parallel, expanding the combination and keeping only first-order terms in the small coefficients, the parallel resistance varies with the same effective coefficient. Both combinations therefore have temperature coefficient

\frac{α _{1} + α _{2}}{2}

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option

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JEE Main · 2010 · 11 Apr 2010 · Offline · Current Electricity

Solution

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