Question

A temperature-stable resistor is made by connecting a resistor made of silicon in series with one made of iron. If the required total resistance is $1000\Omega$in a wide temperature range aroundrole="math" localid="1662722083861" $20°C$, what should be the resistance of the (a) silicon resistor and (b) iron resistor? (See Table 26-1)

Short answer

• a)The resistance of the silicon resistor is $85.0\Omega$.
• b) The resistance of the iron resistor is $915\Omega$.

Step-by-step solution

The given data

• a)Required total resistance,$R=1000\Omega .$
• b)Temperature range,$T=20°C$
• c)The temperature coefficient of silicon material,role="math" localid="1662722513380" ${\alpha }_s=-70\times {10}^{-3}/k$
• d)The temperature coefficient of iron material,${\alpha }_1=6.5\times {10}^{-3}/K$

Understanding the concept of resistivity and resistance

The silicon resistor and the iron resistor are connected in series. Both resistors are temperature-dependent, but we want the combination to be independent of temperature. Thus, using the condition of resistivity of material dependence on temperature, the required resistance values can be calculated considering the direct relation of the resistivity and resistance.

Formula:

The resistivity of the materials dependence on the temperature,

$\rho ={\rho }_0\left[1-\alpha (T-T_0)\right],where\alpha =Temperaturecoeeficientofresistance$ (i)

The resistance of the material,$R=\frac{\rho L}{A}$ (ii)

The equivalent resistance for a series combination,$R_{eq}=\underset{1}{\overset{n}{\sum R_n}}$ (iii)

Calculation of the resistance of the silicon resistor

(a)

From equation (ii), we can see that$R\alpha \rho$$R\alpha \rho$

Now, using the above relation in equation (i), the resistance of the silicon resistor can be given as:
$R_S=R_{0S}\left[1-{\alpha }_S\left(T-T_0\right)\right]............................\left(a\right)$

And, using the above relation in equation (i), the resistance of the iron resistor can be given as:

$R_I=R_{0I}\left[1-{\alpha }_I\left(T-T_0\right)\right].......................................\left(b\right)$

Now, as the resistors are connected in series, using equation (iii), the required total resistance can be given as:

role="math" localid="1662723676849" $R=R_{0s}+R_{0l}.............................................\left(c\right)$

Now adding equation (a) and (b), we can get the resistance relation as follows:

$$\begin{gathered} R\left(T\right)=R_{0S}\left[1-{\alpha }_S\left(T-T_0\right)\right]+R_{0l}\left[1-{\alpha }_l\left(T-T_0\right)\right] \\ =R+\left(R_{0S}{\alpha }_S+R_{0l}{\alpha }_1\right)\left(T-T_0\right) \end{gathered}$$

For the total resistance to be independent of temperature, the above value with term temperature term needs to be zero. Using this condition, the resistance value of the silicon resistor can be given using the given data as follows:

$$\begin{gathered} \left(R_{0S}{\alpha }_3+R_{0I}{\alpha }_1\right)=0 \\ {R}_{0S}{\alpha }_S+\left(R-R_{0S}\right){\alpha }_1=0(\because Fromequation(c),R_{0s}=(R-R_{0s}\left)\right) \\ {R}_{0s}=\frac{R{\alpha }_I}{{\alpha }_I-{\alpha }_S} \\ =\frac{\left(1000\Omega \right)\left(6.5\times {10}^{-3}/K\right)}{6.5\times {10}^{-3}/K-\left(-70.0\times {10}^{-3}/K\right)} \\ =85.0\Omega \end{gathered}$$

Hence, the value of the resistance of silicon is $85.0\Omega$.

 Calculation of the resistance of the iron resistor

(b)

Now, using the given data and above value in equation (c), the resistance of the iron resistor can be given as follows:

$$\begin{gathered} R_{I0}=R-R_{S0} \\ =1000\Omega -85.0\Omega \\ =915\Omega \end{gathered}$$

Hence, the value of the resistance of the iron is $915\Omega$.