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Q.
The $V-I$ graph for a conductor at temperature $T_{1}$ and $T_{2}$ are as shown in the figure. The term $\left(T_{2}-T_{1}\right)$ is proportional to
Current Electricity
Solution:
The slope of $V$ -I curve gives the resistance.
$R_{1}=\tan \theta=R_{0}\left(1+\alpha T_{1}\right)$
and $R_{2}=\cot \theta=R_{0}\left(1+\alpha T_{2}\right)$
$\cot \theta-\tan \theta=R_{0}\left(1+\alpha T_{2}\right)-R_{0}\left(1+\alpha T_{1}\right)=R_{0} \alpha\left(T_{2}-T_{1}\right)$
or $T_{2}-T_{1}=\frac{1}{\alpha R_{0}}(\cot \theta-\tan \theta)$
$=\frac{1}{\alpha R_{0}}\left(\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}\right)$
$=\frac{2 \cos 2 \theta}{\alpha R_{0} \sin 2 \theta}=\frac{2}{\alpha R_{0}} \cot 2 \theta$
or $T_{2}-T_{1} \propto \cot 2 \theta$
