Chapter 2, Problem 2.70HP

To determine

(a)

To plot:

A graph of $R_{\text{th}}\left(T\right)$ as a function of the surrounding temperature $T$ .

Answer to Problem 2.70HP

The plot of $R_{\text{th}}\left(T\right)$ as a function of the surrounding temperature $T$ is shown in Figure 1.

Explanation of Solution

Calculation:

The formula to calculate the resistance $R_{\text{th}}$ at temperature $T$ is given by,

  $R_{\text{th}}\left(T\right)=R_0{e}^{-\beta \left(T-T_0\right)}$ ........(1)

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $350\text{K}$ for $T$ in the above equation.

  $\begin{array}{c}{R}_{\text{th}}\left(350\text{K}\right)=\left(300\Omega \right)e^{-\left[-0.01{\text{K}}^{-1}\left(350\text{K}-298\text{K}\right)\right]}\\ =504.6\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $400\text{K}$ for $T$ in equation (1)

  $\begin{array}{c}{R}_{\text{th}}\left(400\text{K}\right)=\left(300\Omega \right)e^{-\left[-0.01{\text{K}}^{-1}\left(400\text{K}-298\text{K}\right)\right]}\\ =831.95\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $450\text{K}$ for $T$ in equation (1)

  $\begin{array}{c}{R}_{\text{th}}\left(450\text{K}\right)=\left(300\Omega \right)e^{-\left[-0.01{\text{K}}^{-1}\left(450\text{K}-450\text{K}\right)\right]}\\ =1371.67\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $500\text{K}$ for $T$ in equation (1)

  $\begin{array}{c}{R}_{\text{th}}\left(500\text{K}\right)=\left(300\Omega \right)e^{-\left[-0.01{\text{K}}^{-1}\left(500\text{K}-450\text{K}\right)\right]}\\ =2261.5\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $550\text{K}$ for $T$ in equation (1)

  $\begin{array}{c}{R}_{\text{th}}\left(550\text{K}\right)=\left(300\Omega \right)e^{-\left[-0.01{\text{K}}^{-1}\left(550\text{K}-450\text{K}\right)\right]}\\ =3728.57\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $600\text{K}$ for $T$ in equation (1)

  $\begin{array}{c}{R}_{\text{th}}\left(600\text{K}\right)=\left(300\Omega \right)e^{-\left[-0.01{\text{K}}^{-1}\left(600\text{K}-450\text{K}\right)\right]}\\ =10135.32\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $650\text{K}$ for $T$ in equation (1)

  $\begin{array}{c}{R}_{\text{th}}\left(650\text{K}\right)=\left(300\Omega \right)e^{-\left[-0.01{\text{K}}^{-1}\left(650\text{K}-450\text{K}\right)\right]}\\ =16710.33\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $700\text{K}$ for $T$ in equation (1)

  $\begin{array}{c}{R}_{\text{th}}\left(700\text{K}\right)=\left(300\Omega \right)e^{-\left[-0.01{\text{K}}^{-1}\left(700\text{K}-450\text{K}\right)\right]}\\ =27550.68\Omega \end{array}$

The plot of $R_{\text{th}}\left(T\right)$ as a function of the surrounding temperature $T$ for $350\le T\le 750$ is shown below.

The required diagram is shown in Figure 1

  

To determine

(b)

The expression for the equivalent resistance.

To plot:

A sketch of resistance $R_{\text{th}}\left(T\right)$ on the same graph for part a.

Answer to Problem 2.70HP

The expression for the equivalent resistance is $R_{\text{eq}}=\frac{R_{th}\left(T\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(T\right)}$ and the plot for the equivalent resistance for different temperature ranges is shown in Figure 2.

Explanation of Solution

Calculation:

The expression for the equivalent resistance is given by,

  $R_{\text{eq}}=\frac{R_{th}\left(T\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(T\right)}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $350\text{K}$ for $T$ in the above equation.

  $R_{\text{eq}}=\frac{R_{th}\left(350\text{K}\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(350\text{K}\right)}$

Substitute $504.6\Omega$ for $R_{th}\left(350\text{K}\right)$ in the above equation.

  $\begin{array}{c}{R}_{\text{eq}}=\frac{504.6\Omega \left(250\Omega \right)}{250\Omega +504.6\Omega }\\ =167.17\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $400\text{K}$ for $T$ in equation (1)

  $R_{\text{eq}}=\frac{R_{th}\left(400\text{K}\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(400\text{K}\right)}$

Substitute $831.95\Omega$ for $R_{th}\left(400\text{K}\right)$ in the above equation.

  $\begin{array}{c}{R}_{\text{eq}}=\frac{831.95\Omega \left(250\Omega \right)}{250\Omega +831.95\Omega }\\ =192.23\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $450\text{K}$ for $T$ in equation (1)

  $R_{\text{eq}}=\frac{R_{th}\left(450\text{K}\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(450\text{K}\right)}$

Substitute $1371.67\Omega$ for $R_{th}\left(450\text{K}\right)$ in the above equation.

  $\begin{array}{c}{R}_{\text{eq}}=\frac{1371.67\Omega \left(250\Omega \right)}{250\Omega +1371.67\Omega }\\ =211.46\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $500\text{K}$ for $T$ in equation (1)

  $R_{\text{eq}}=\frac{R_{th}\left(500\text{K}\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(500\text{K}\right)}$

Substitute $2261.5\Omega$ for $R_{th}\left(500\text{K}\right)$ in the above equation.

  $\begin{array}{c}{R}_{\text{eq}}=\frac{2261.5\Omega \left(250\Omega \right)}{250\Omega +2261.5\Omega }\\ =225.11\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $550\text{K}$ for $T$ in equation (1)

  $R_{\text{eq}}=\frac{R_{th}\left(550\text{K}\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(550\text{K}\right)}$

Substitute $3728.57\Omega$ for $R_{th}\left(550\text{K}\right)$ in the above equation.

  $\begin{array}{c}{R}_{\text{eq}}=\frac{3728.57\Omega \left(250\Omega \right)}{250\Omega +3728.57\Omega }\\ =234.30\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $600\text{K}$ for $T$ in equation (1)

  $R_{\text{eq}}=\frac{R_{th}\left(600\text{K}\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(600\text{K}\right)}$

Substitute $6147.38\Omega$ for $R_{th}\left(600\text{K}\right)$ in the above equation.

  $\begin{array}{c}{R}_{\text{eq}}=\frac{6147.38\Omega \left(250\Omega \right)}{250\Omega +6147.38\Omega }\\ =240.23\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $650\text{K}$ for $T$ in equation (1)

  $R_{\text{eq}}=\frac{R_{th}\left(650\text{K}\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(650\text{K}\right)}$

Substitute $10135.32\Omega$ for $R_{th}\left(650\text{K}\right)$ in the above equation.

  $\begin{array}{c}{R}_{\text{eq}}=\frac{10135.32\Omega \left(250\Omega \right)}{250\Omega +10135.32\Omega }\\ =244\Omega \end{array}$

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $700\text{K}$ for $T$ in equation (1)

  $R_{\text{eq}}=\frac{R_{th}\left(700\text{K}\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(700\text{K}\right)}$

Substitute $16710.33\Omega$ for $R_{th}\left(700\text{K}\right)$ in the above equation.

  $\begin{array}{c}{R}_{\text{eq}}=\frac{16710.33\Omega \left(250\Omega \right)}{250\Omega +16710.33\Omega }\\ =246.31\Omega \end{array}$ s

Substitute $300\Omega$ for $R_0$, $-0.01{\text{K}}^{-1}$ for $\beta$, $298\text{K}$ for $T_0$ and $750\text{K}$ for $T$ in equation (1)

  $R_{\text{eq}}=\frac{R_{th}\left(750\text{K}\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(750\text{K}\right)}$

Substitute $27550.68\Omega$ for $R_{th}\left(750\text{K}\right)$ in the above equation.

  $\begin{array}{c}{R}_{\text{eq}}=\frac{27550.68\Omega \left(250\Omega \right)}{250\Omega +27550.68\Omega }\\ =247.75\Omega \end{array}$ s

The plot for the equivalent resistance for different temperature ranges is shown in Figure 2

  

Conclusion:

Therefore, the expression for the equivalent resistance is $R_{\text{eq}}=\frac{R_{th}\left(T\right)\left(250\Omega \right)}{250\Omega +R_{th}\left(T\right)}$ and the plot for the equivalent resistance for different temperature ranges is shown in Figure 2.