91影视

Write the expression for the Ohm鈥檚 law.

$\mathrm{V}=\mathrm{IR}$

Here, I is current in ampere$\left(\mathrm{A}\right)$,$R$is resistance in ohms$\left(\mathrm{惟}\right)$and$V$is the potential difference volt$\left(\mathrm{V}\right)$.

Write the expression for the resistance in terms of the voltage and resistivity.

$$\begin{gathered} R=\frac{V}{I} \\ =\frac{蚁L}{A} \end{gathered}$$

Here,$蚁$is resistivity$\mathrm{惟}.\mathrm{m}$, L is length in m and A is area in${\mathrm{m}}^2$.

Write$\mathit{蚁}$of the material is the ratio of the magnitude of the electric field$\mathit{E}$and

current density J .

$$\begin{gathered} 蚁=\frac{E}{J} \\ 蚁\left(T\right)={蚁}_0\left[1+伪\left(T-T_0\right)\right] \end{gathered}$$

Write the expression for the temperature coefficient as:

$伪=\frac{\left({\displaystyle \frac{R_T}{R_0}}\right)-1}{\left(T-T_0\right)}$

(a)

Consider the given parameters:

$$\begin{gathered} \mathrm{L}=1.50\mathrm{m} \\ \mathrm{d}=0.500\mathrm{cm} \\ \mathrm{V}=15\mathrm{V} \\ {\mathrm{I}}_0=18.50\mathrm{A} \end{gathered}$$

The resistivity of cylindrical rod is:

$蚁=\frac{R_{0}A}{L}$

Substitute the value and solve as,

$$\begin{gathered} 蚁=\frac{\left({\displaystyle \frac{15.0}{18.50}}\right)\left\{\mathrm{蟺}{\left({\displaystyle \frac{0.500}{2}}\right)}^2\right\}}{1.50}\left[鈭�R_0=\frac{V}{I_0}\right] \\ =\frac{\left(0.811\right)\mathrm{蟺}\left(0.0025\right)}{1.50} \\ =1.60脳{10}^{-5}\mathrm{惟}.\mathrm{m} \end{gathered}$$

Hence, the resistivity at ${20}^{鈭�}\mathrm{C}$is $1.60脳{10}^{-5}\mathrm{惟}.\mathrm{m}$.

(b)

Consider the given parameters:

$$\begin{gathered} T_0=20.0^{鈭�}C \\ T=92.0^{鈭�}C \\ I=17.2A \end{gathered}$$

The temperature coefficient $伪$is:

$伪=\frac{\left({\displaystyle \frac{R_T}{R_0}}\right)-1}{\left(T-T_0\right)}$

Here, $R_T=\frac{V}{I}$and $R_0$is 0.811 role="math" localid="1655788149360" $\mathrm{惟}$.

Substitute the value and solve.

$$\begin{gathered} 伪=\frac{\left({\displaystyle \frac{15/17.2}{0.811}}\right)-1}{\left(92-20\right)} \\ =\frac{\left(0.872/0.811\right)-1}{72} \\ =0.00105{\left({\mathrm{C}}^{鈭�}\right)}^{-1} \end{gathered}$$

Hence, temperature coefficient of resistivity at ${20}^{鈭�}C$for the material of the rod is

0.00105${\left({\mathrm{C}}^{鈭�}\right)}^{-1}$.