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 :

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: