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  • V-I graph for a conductor (Platinum wire) at temperatures and <img alt="T_2" src="https://entrancecorner.oncodecogs.com/g

V-I graph for a conductor (Platinum wire) at temperatures T_1 and T_2 is as shown \left(T_2-T_1\right) is proportional to

Option: 1

\cos 2 \theta\, \, \, \,


Option: 2

\sin 2 \, \theta


Option: 3

\cot 2\, \theta


Option: 4

\tan 2 \, \theta


Answers (1)

best_answer

We know that the slope of V-l graph gives resistance.

Therefore, resistance at T_1=\tan \theta

Resistance at T_2=\tan (90-\theta)=\cot \theta

\mathrm{R}_{\mathrm{T}_2}-\mathrm{R}_{\mathrm{T}_1}=\cot \theta-\tan \theta

=\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}=\frac{\cos ^2 \theta-\sin ^2 \theta}{\sin \theta \cos \theta}

\mathrm{R}_{\mathrm{T}_2}-\mathrm{R}_{\mathrm{T}_1}=\frac{2 \cos 2 \theta}{\sin 2 \theta}=2 \cot 2 \theta

\text { Now } \mathrm{R}_{\mathrm{T}_2}=\mathrm{R}_{\mathrm{t}_2}=\mathrm{R}_0\left(1+\propto \mathrm{t}_2\right)

R_{T_1}=R_{t_2}=R_0\left(1+\propto t_1\right)

where \alpha \ is temperature coefficient of resistance 

\mathrm{R}_{\mathrm{T}_2}-\mathrm{R}_{\mathrm{T}_1}=\mathrm{R}_{\mathrm{t}_2}-\mathrm{R}_{\mathrm{t}_1}=\mathrm{R}_0\left[1+\alpha \mathrm{t}_2-1-\alpha \mathrm{t}_1\right]=\mathrm{R}_0 \alpha\left(\mathrm{t}_2-\mathrm{t}_1\right)=\mathrm{R}_0 \alpha\left(\mathrm{T}_2-\mathrm{T}_1\right)

\therefore \quad R_{T_2}-R_{T_1}=R_0 \alpha\left[T_2-T_1\right]=2 \cot 2 \theta

\therefore \quad T_2-T_1=\frac{2}{R_0 \alpha} \cot 2 \theta \text { or } \quad T_2-T_1 \propto \cot 2 \theta

Posted by

Devendra Khairwa

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