The $V-I$ graph for a conductor at temperature $T_1$ and $T_2$ are as shown in the figure. $T_2-T_1$ is proportional to

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Aicharya · Accepted Answer

$V-I$ graph for conductor at temperature $T_1$ and $T_2$ is shown in the figure. We know that resistance of a conductor is directly proportional to its temperature. \begin{verbatim} $\begin{aligned} & \text{Hence, } R_1 \propto T_1 \Rightarrow \tan \theta \propto T_1 \ & \Rightarrow \tan \theta=K T_1 \quad \ldots (i) \ & \text{and } R_2 \propto T_2 \ & \Rightarrow \tan \left(90^{\circ}-\theta\right) \propto T_2 \ & \Rightarrow \cot \theta \propto T_2 \ & \Rightarrow \cot \theta=K T_2 \quad \ldots (ii) \end{aligned}$ $\therefore$ From Eqs. (i) and (ii), we get $\begin{aligned} & K T_2-K T_1=\cot \theta-\tan \theta \ & K\left(T_2-T_1\right)=\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta} \ & \Rightarrow T_2-T_1=\frac{\cos ^2 \theta-\sin ^2 \theta}{K \sin \theta \cos \theta} \ & \Rightarrow T_2-T_1=\frac{\cos 2 \theta}{\frac{1}{2} \sin 2 \theta} \Rightarrow T_2-T_1=\frac{2 \cot 2 \theta}{K} \ & \Rightarrow T_2-T_1 \propto \cot 2 \theta \ \end{aligned}$

The \(V-I\) graph for a conductor at temperature \(T_1\) and \(T_2\) are as shown in the figure. \(T_2-T_1\)…

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