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

Vedclass · Accepted Answer

$\cot 2	heta$

Vedclass · Answer

(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,$(T_2 - T_1) \propto \cot 2	heta$.

$I (A)$	$V (V)$	$I (A)$	$V (V)$
$0.2$	$3.94$	$3.0$	$59.2$
$0.4$	$7.87$	$4.0$	$78.8$
$0.6$	$11.8$	$5.0$	$98.6$
$0.8$	$15.7$	$6.0$	$118.5$
$1.0$	$19.7$	$7.0$	$138.2$
$2.0$	$39.4$	$8.0$	$158.0$

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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What conclusion can you draw from the following observations on a resistor made of alloy manganin?
$I (A)$ $V (V)$ $I (A)$ $V (V)$
$0.2$ $3.94$ $3.0$ $59.2$
$0.4$ $7.87$ $4.0$ $78.8$
$0.6$ $11.8$ $5.0$ $98.6$
$0.8$ $15.7$ $6.0$ $118.5$
$1.0$ $19.7$ $7.0$ $138.2$
$2.0$ $39.4$ $8.0$ $158.0$

Which of the following graphs represents an ohmic resistance?

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