Question

Two resistance coils connected in series have resistance of $600\Omega$ and $300\Omega$ at ${20}^{\circ }C$ and temperature co-efficient of resistances are ${0.001}^{\circ }C$ and ${0.004}^{\circ }C$ respectively. What is the effective temperature coefficient of the combination?

• A. $0.001{}^{\circ }{C}^{-1}$
• B. $0.002{}^{\circ }{C}^{-1}$
• C. $0.003{}^{\circ }{C}^{-1}$
• D. $0.004{}^{\circ }{C}^{-1}$

Answer 1

The correct option is B $0.002{}^{\circ }{C}^{-1}$

In series combination the equivalent temperature coefficient of resistance is given as,$${\alpha }_{eq}=\frac{R_1{\alpha }_1+R_2{\alpha }_2}{R_1+R_2}$$Substituting the values,

$\Rightarrow {\alpha }_{eq}=\frac{\left(600\right)\left(0.001\right)+\left(300\right)\left(0.004\right)}{600+300}$

$\Rightarrow {\alpha }_{eq}=0.002{}^{\circ }{C}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (b) is the correct answer.

Why this question ? When two resistors are connected in parallel, effective thermal co-efficient of resistance of the combination is, $({\alpha }_{eq}{)}_{series}=\frac{R_1{\alpha }_1+R_2{\alpha }_2}{R_1+R_2}$ When two resistors are connected in series, effective thermal co-efficient of resistance of the combination is, $({\alpha }_{eq}{)}_{parallel}=\frac{R_1{\alpha }_2+R_2{\alpha }_1}{R_1+R_2}$