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

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(C) For a conductor,the resistance $R$ is directly proportional to the temperature $T$,i.e.,$R \propto T$. In a $V-i$ graph,the slope represents the r

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$\cot 2	heta$

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(C) For a conductor,the resistance $R$ is directly proportional to the temperature $T$,i.e.,$R \propto T$. In a $V-i$ graph,the slope represents the resistance $R = \frac{V}{i}$. From the figure,for temperature $T_1$,the slope is $	an 	heta$. Thus,$R_1 = 	an 	heta \propto T_1$,which gives $T_1 = k 	an 	heta$ for some constant $k$. For temperature $T_2$,the angle with the $V$-axis is $	heta$,so the angle with the $i$-axis is $(90^\circ - 	heta)$. Thus,$R_2 = 	an(90^\circ - 	heta) = \cot 	heta \propto T_2$,which gives $T_2 = k \cot 	heta$. Now,$(T_2 - T_1) = k(\cot 	heta - 	an 	heta)$. Using trigonometric identities: $(T_2 - T_1) = k \left( \frac{\cos 	heta}{\sin 	heta} - \frac{\sin 	heta}{\cos 	heta} ight) = k \left( \frac{\cos^2 	heta - \sin^2 	heta}{\sin 	heta \cos 	heta} ight) = k \left( \frac{\cos 2	heta}{\frac{1}{2} \sin 2	heta} ight) = 2k \cot 2	heta$. Therefore,$(T_2 - T_1) \propto \cot 2	heta$.

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

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