Question

Two wires of resistance R₁ and R₂ have temperature co-efficient of resistance α₁ and α₂ respectively. These are joined in series. The effective temperature co-efficient of resistance is

• A. $\frac{\alpha_{1} + \alpha_{2}}{2}$
• B. $\sqrt{\alpha_{1}\alpha_{2}}$
• C. $\frac{\alpha_{1}R_{1} + \alpha_{2}R_{2}}{R_{1} + R_{2}}$
• D. $\frac{\sqrt{R_{1}R_{2}\alpha_{1}\alpha_{2}}}{\sqrt{R_{1}^{2} + R_{2}^{2}}}$

Answer 1

Answer: (C)

$\frac{\alpha_{1}R_{1} + \alpha_{2}R_{2}}{R_{1} + R_{2}}$

Answer 2

Suppose at t^oC resistances of the two wires becomes $R_{1t}$ and $R_{2t}$ respectively and equivalent resistance becomes R_t. In series grouping R_t = R_1t + R_2t, also R_1t = R₁(1 + α₁t) and R_2t = R₂(1 + α₂t)

R_t = R₁(1 + α₁t) + R₂(1 + α₂t) = (R₁ + R₂) + (R₁α₁ + R₂α₂) t = $(R_{1} + R_{2})\left\lbrack 1 + \frac{R_{1}\alpha_{1} + R_{2}\alpha_{2}}{R_{1} + R_{2}}t \right\rbrack$.

Hence effective temperature co-efficient is $\frac{R_{1}\alpha_{1} + R_{2}\alpha_{2}}{R_{1} + R_{2}}$.