The V-I graph for a conductor at temperatures | vedclass

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(C) The $V-I$ graph for a conductor at temperatures $T_1$ and $T_2$ is shown in the figure.We know that the resistance $R$ of a conductor is directly

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$\cot 2 	heta$

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(C) The $V-I$ graph for a conductor at temperatures $T_1$ and $T_2$ is shown in the figure.We know that the resistance $R$ of a conductor is directly proportional to its temperature $T$ (i.e.,$R \propto T$).For the $V-I$ graph,the slope represents the resistance $R = \frac{V}{I}$.From the figure,the slope of the line for $T_1$ with respect to the $I$-axis is $	an 	heta$. Thus,$R_1 \propto 	an 	heta \Rightarrow R_1 = K 	an 	heta$,where $K$ is a constant.Similarly,the slope of the line for $T_2$ with respect to the $I$-axis is $	an(90^{\circ}-	heta) = \cot 	heta$. Thus,$R_2 \propto \cot 	heta \Rightarrow R_2 = K \cot 	heta$.Therefore,$T_2 - T_1 \propto R_2 - R_1 = K(\cot 	heta - 	an 	heta)$.$T_2 - T_1 \propto K \left( \frac{\cos 	heta}{\sin 	heta} - \frac{\sin 	heta}{\cos 	heta} ight) = K \left( \frac{\cos^2 	heta - \sin^2 	heta}{\sin 	heta \cos 	heta} ight)$.Using trigonometric identities $\cos^2 	heta - \sin^2 	heta = \cos 2 	heta$ and $\sin 	heta \cos 	heta = \frac{1}{2} \sin 2 	heta$,we get:$T_2 - T_1 \propto K \left( \frac{\cos 2 	heta}{\frac{1}{2} \sin 2 	heta} ight) = 2K \cot 2 	heta$.Thus,$T_2 - T_1 \propto \cot 2 	heta$.

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

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