Question

Two resistance with temperature coefficients of resistance ${\alpha }_1$ and ${\alpha }_2$ have resistances $R_{01}$ and $R_{02}$ at $0^{\circ }$C. Find the temperature coefficient of the compound resistor consisting of the two resistors connected in series.

• A. $\alpha =\frac{R_{01}{\alpha }_2+R_{02}{\alpha }_1}{R_{01}+R_{02}}$
• B. $\alpha =\frac{R_{01}{\alpha }_1+R_{02}{\alpha }_2}{R_{01}+R_{02}}$
• C. $\alpha =\frac{R_{01}+R_{02}}{R_{01}{\alpha }_1+R_{02}{\alpha }_2}$
• D. $\alpha =\frac{R_{01}+R_{02}}{R_{01}{\alpha }_2+R_{02}{\alpha }_1}$

Answer 1

The correct option is B $\alpha =\frac{R_{01}{\alpha }_1+R_{02}{\alpha }_2}{R_{01}+R_{02}}$



at $0^{\circ }$C $R_{01}$ and $R_{02}$ are the resistance and the combined resistance will be
$\begin{array}{}{R}_0=R_{01}+R_{02}& \text{(I)}\end{array}$

At $t^{\circ }$C $R_{01}$ will become $R_{01}(1+{\alpha }_{1}t)$
Similarly, $t^{\circ }$C $R_{02}$ will become $R_{02}(1+{\alpha }_{2}t)$
and the overall resistance will be $R_0(1+\alpha t)$
substitute these resistance at temperature in equation (I) we get
$R_{01}(1+{\alpha }_{1}t)$+$R_{02}(1+{\alpha }_{2}t)$=$R_0(1+\alpha t)$

$R_{01}+R_{01}{\alpha }_{1}t+R_{02}+R_{02}{\alpha }_{2}t$
$=R_0+R_0\alpha t$

$R_{01}+R_{01}{\alpha }_{1}t+R_{02}+R_{02}{\alpha }_{2}t$
$=R_{01}+R_{02}+(R_{01}+R_{02})\alpha t$

$\alpha =\frac{R_{01}{\alpha }_1+R_{02}{\alpha }_2}{R_{01}+R_{02}}$