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Question: The V-i graph for a conductor at temperature T<sub>1</sub> and T<sub>2</sub> are as shown in the fig...

The V-i graph for a conductor at temperature T1 and T2 are as shown in the figure. (T2 – T1) is proportional to

A

cos 2θ

B

sinθ

C

cot 2θ

D

tanθ

Answer

cot 2θ

Explanation

Solution

As we know, for conductors resistance ∝ Temperature.

From figure R1 ∝ T1 ⇒ tanθ ∝ T1 ⇒ tanθ = kT1 …. (i)

(k = constant)

and R2 ∝ T2 ⇒ tan (90o – θ) ∝ T2 ⇒ cotθ = kT2 ……..(ii)

From equation (i) and (ii) k(T2−T1)=(cot⁡θ−tan⁡θ)k(T_{2} - T_{1}) = (\cot\theta - \tan\theta)

(T2−T1)=(cos⁡θsin⁡θ−sin⁡θcos⁡θ)=(cos⁡2θ−sin⁡2θ)sin⁡θcos⁡θ=cos⁡2θsin⁡θcos⁡θ=2cot⁡2θ(T_{2} - T_{1}) = \left( \frac{\cos\theta}{\sin\theta} - \frac{\sin\theta}{\cos\theta} \right) = \frac{(\cos^{2}\theta - \sin^{2}\theta)}{\sin\theta\cos\theta} = \frac{\cos 2\theta}{\sin\theta\cos\theta} = 2\cot 2\theta ⇒ (T2 – T1) ∝ cot 2θ