The V-i graphs for a conductor at temperature | vedclass

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(C) For a conductor,resistance $R \propto T$. From the $V-i$ graph,the slope represents resistance $R = \frac{V}{i} = 	an \phi$,where $\phi$ is the a

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$\cot \, 2	heta$

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(C) For a conductor,resistance $R \propto T$. From the $V-i$ graph,the slope represents resistance $R = \frac{V}{i} = 	an \phi$,where $\phi$ is the angle with the $i$-axis.For temperature $T_1$,the angle with the $i$-axis is $	heta$. Thus,$R_1 = 	an 	heta = k T_1$ (where $k$ is a constant). $(i)$For temperature $T_2$,the angle with the $V$-axis is $	heta$,so the angle with the $i$-axis is $(90^\circ - 	heta)$. Thus,$R_2 = 	an(90^\circ - 	heta) = \cot 	heta = k T_2$. (ii)Subtracting $(i)$ from (ii): $k(T_2 - T_1) = \cot 	heta - 	an 	heta$.$(T_2 - T_1) \propto \left( \frac{\cos 	heta}{\sin 	heta} - \frac{\sin 	heta}{\cos 	heta} ight) = \frac{\cos^2 	heta - \sin^2 	heta}{\sin 	heta \cos 	heta} = \frac{\cos 2	heta}{\frac{1}{2} \sin 2	heta} = 2 \cot 2	heta$.Therefore,$(T_2 - T_1) \propto \cot 2	heta$.

The $V-i$ graphs for a conductor at temperatures $T_1$ and $T_2$ are shown. $(T_2 - T_1)$ is proportional to:

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