First slide

Resistors

Question

The V-I graph for a conductor temperatures $T_{1} {and T}_{2}$ are as shown in the fig.
$(T_{2} - T_{1})$ is proportional to

Moderate

A

$cos 2θ$
B

$sin θ$
C

$cot 2θ$
D

$tan θ$

Solution

The slope of V – I graph gives resistance
$R_{1} {= tan θ , R}_{2} = tan (90-θ) = cot θ L e t s a y α i s t e m p e r a t u r e c o e f f i c i e n t R_{1} {=R}_{0} (1 + {αT}_{1}) {, R}_{2} {=R}_{0} (1 + {αT}_{2}) \Rightarrow (R_{2} {- R}_{1}) = R_{0} α (T_{2} - T_{1}) \Rightarrow cot θ - tan θ = α (T_{2} {-T}_{1}) \Rightarrow (T_{2} {-T}_{1}) \propto \frac{2 [{cos}^{2} {θ-sin}^{2} θ]}{2cos θsinθ} \Rightarrow (T_{2} {-T}_{1}) \propto \frac{2cos2θ}{sin2θ} \Rightarrow (T_{2} {-T}_{1}) \propto cot (2 θ)$

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