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

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(C) In a $V-I$ graph,the slope represents the resistance $R$. For temperature $T_1$,the line makes an angle $	heta$ with the $I$-axis. Thus,$R_1 =

Vedclass · Accepted Answer

$\cot 2	heta$

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(C) In a $V-I$ graph,the slope represents the resistance $R$. For temperature $T_1$,the line makes an angle $	heta$ with the $I$-axis. Thus,$R_1 = 	an 	heta$. Since $R \propto T$,we have $R_1 = k T_1 = 	an 	heta$. For temperature $T_2$,the line makes an angle $	heta$ with the $V$-axis. The angle with the $I$-axis is $(90^\circ - 	heta)$. Thus,$R_2 = 	an(90^\circ - 	heta) = \cot 	heta$. Since $R \propto T$,we have $R_2 = k T_2 = \cot 	heta$. Now,consider $(T_2 - T_1) \propto (R_2 - R_1)$. $(T_2 - T_1) \propto (\cot 	heta - 	an 	heta)$. Using trigonometric identities: $\cot 	heta - 	an 	heta = \frac{\cos 	heta}{\sin 	heta} - \frac{\sin 	heta}{\cos 	heta} = \frac{\cos^2 	heta - \sin^2 	heta}{\sin 	heta \cos 	heta} = \frac{\cos 2	heta}{\frac{1}{2} \sin 2	heta} = 2 \cot 2	heta$. Therefore,$(T_2 - T_1) \propto \cot 2	heta$.

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

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