Question 345: Current Electricity

Question

In an experiment, the resistance of a material
is plotted as a function of temperature (in some
range). As shown in the figure, it is a straight
line. One may conclude that :
![](https://imagex.cdn.examgoal.net/3rsa0w2w9jwyuks8r/670d173e-2149-4f8c-8153-985fee81964b/30c19cb0-9027-11e9-86c1-73f7f991b766/file-3rsa0w2w9jwyuks8s.png)

Accepted Answer

${{{1 \over {{T^2}}}} \over {{1 \over {T_0^2}}}} + {{\ln \,R\left( T \right)} \over {\ln \,R\left( {{T_o}} \right)}} = 1$
$ \Rightarrow $ ln R(T) = ln R(T
o
) $\left( {1 - {{T_o^2} \over {{T^2}}}} \right)$
$R(T) = {R_o}{e^{ - \left( {{{T_o^2} \over {{T^2}}}} \right)}}$

Answer

$R(T) = {R_0}{e^{ - {T^2}/T_0^2}}$

Answer

$R(T) = {{{R_0}} \over {{T^2}}}$

Answer

$R(T) = {R_0}{e^{ {T^2}/T_0^2}}$

Answer

$R(T) = {R_0}{e^{ - T_0^2/{T^2}}}$

Current Electricity

In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line. One may conclude that :

Choose the correct answer: