Irodov 3.149: Resistances Weigh Series, Conductances Weigh Parallel

Problem Irodov 3.149

Two conductors, $1$ and $2$ , have temperature coefficients of resistance $\alpha_1$ and $\alpha_2$ . At $0\,^\circ$ C the resistance of conductor $2$ is $\eta$ times that of conductor $1$ . Find the temperature coefficient of resistance of the segment formed by the two conductors when they are joined

in series;
in parallel.

(a) Conductors 1 (R_1) and 2 (R_2) in series. — (a) Conductors $1$ ( $R_1$ ) and $2$ ( $R_2$ ) in series.

(b) The same conductors, R_1 and R_2, in parallel. — (b) The same conductors, $R_1$ and $R_2$ , in parallel.

Given

conductor $2$ : $\eta$ times the resistance of conductor $1$ at $0\,^\circ$ C
coefficients $\alpha_1$ , $\alpha_2$
small temperature changes (assumed)

Find

the coefficient

\alpha

of (a) the series, (b) the parallel segment

Solution

Step 1What a temperature coefficient means

A conductor’s resistance at the temperature $t$ (in $^\circ$ C) grows linearly from its value at $0\,^\circ$ C, with the coefficient as the relative rate. For the two conductors, with $R_{10}$ and $R_{20} = \eta R_{10}$ their resistances at $0\,^\circ$ C:

\begin{gathered} R_1 = R_{10}\,(1 + \alpha_1 t)\\ R_2 = \eta R_{10}\,(1 + \alpha_2 t) \end{gathered} \tag{1}

The coefficient $\alpha$ of the segment is the number that puts its resistance in the same form, $R = R_0\,(1 + \alpha t)$ , with $R_0$ its own resistance at $0\,^\circ$ C.

Step 2(a) Series: the resistances add

Adding the two lines of (1) and taking out the resistance at $0\,^\circ$ C, $R_0 = (1 + \eta)R_{10}$ :

\begin{aligned} R &= R_{10}\,(1 + \alpha_1 t) + \eta R_{10}\,(1 + \alpha_2 t)\\ &= (1 + \eta)R_{10}\left(1 + \frac{\alpha_1 + \eta\alpha_2}{1 + \eta}\,t\right) \end{aligned}

This is exactly linear in $t$ , so the coefficient is read off with no approximation:

\alpha = \frac{\alpha_1 + \eta\alpha_2}{1 + \eta}

(2)

Step 3(b) Parallel: the conductances add

In parallel the reciprocals add. Each reciprocal of (1) is expanded to first order, $1/(1 + \alpha_i t) \approx 1 - \alpha_i t$ , which holds for small temperature changes, $\alpha_i t \ll 1$ (assumed):

\begin{aligned} \frac{1}{R} &= \frac{1}{R_{10}\,(1 + \alpha_1 t)} + \frac{1}{\eta R_{10}\,(1 + \alpha_2 t)}\\ &\approx \frac{1}{R_{10}}\left(1 - \alpha_1 t + \frac{1 - \alpha_2 t}{\eta}\right)\\ &= \frac{1 + \eta}{\eta R_{10}}\left(1 - \frac{\eta\alpha_1 + \alpha_2}{1 + \eta}\,t\right) \end{aligned}

Inverting with the same first-order rule, the resistance at $0\,^\circ$ C is $R_0 = \eta R_{10}/(1 + \eta)$ and

R \approx \frac{\eta R_{10}}{1 + \eta}\left(1 + \frac{\eta\alpha_1 + \alpha_2}{1 + \eta}\,t\right)

so the coefficient of the parallel segment is

\alpha \approx \frac{\alpha_2 + \eta\alpha_1}{1 + \eta}

(3)

The exact parallel resistance is not linear in $t$ , so this coefficient is its slope at $0\,^\circ$ C relative to $R_0$ , valid while $\alpha_i t \ll 1$ .

Check

Equal coefficients, $\alpha_1 = \alpha_2$ : both (2) and (3) give that common value, as they must, since both resistances then scale by the same factor. Extreme ratio $\eta \to 0$ (conductor $2$ negligible): in series (2) gives $\alpha_1$ , since conductor $1$ is almost the whole resistance; in parallel (3) gives $\alpha_2$ , since conductor $2$ shorts conductor $1$ and carries almost all the current.

Remark

The series coefficient is the average of $\alpha_1$ , $\alpha_2$ weighted by the resistances ( $1$ and $\eta$ ), the parallel one the average weighted by the conductances ( $1$ and $1/\eta$ ). This is how temperature-stable resistors are made: a material with a negative coefficient, such as carbon, in series with a metal of positive coefficient, in the ratio $\eta = -\alpha_1/\alpha_2$ , gives a segment whose resistance does not change to first order.