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The current-voltage graphs for a given metallic wire at two different temperatures, $T_1$ and $T_2$ are shown in the figure. The correct relation of temperatures and resistances will be:
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$R_{T_2}=R_{T_1},T_1>T_2$ $R_{T_2}<R_{T_1},T_1>T_2$ $R_{T_2}>R_{T_1},T_2>T_1$ $R_{T_2}<R_{T_1},T_2<T_2$ |
$R_{T_2}>R_{T_1},T_2>T_1$ |
The correct answer is Option (3): $R_{T_2}>R_{T_1},T_2>T_1$ According to Ohm's Law, V = IR, which can be rearranged as I = (1/R)V. This equation is in the form y = mx, where the slope m = 1/R. Therefore, the resistance R is inversely proportional to the slope of the I–V graph:
Evaluating the Slopes: From the graph, the slope for T₁ is steeper than the slope for T₂. Since the slope at T₁ > slope at T₂, it follows that Rₜ₁ < Rₜ₂, or Rₜ₂ > Rₜ₁. Temperature Dependence: For a metallic conductor, the resistance increases as the temperature increases. Since Rₜ₂ > Rₜ₁, it logically follows that the temperature T₂ > T₁. |
