Question

The resistance of a conductor is $5$ ohm at ${{50}^{\text{o}}}\text{C}$ and $\text{6}$ohm at ${{100}^{\text{o}}}\text{C}$. Find its resistance at ${{0}^{\text{o}}}\text{C}$.

Answer 1

When the temperature of a conductor is raised, its electrical resistance increases. Hence, the temperature of the conductor at ${{0}^{\text{o}}}\text{C}$ will be less than its resistance at ${{50}^{\text{o}}}\text{C}$.

Formula used: The resistance of a conductor ${{R}_{t}}$ at ${{t}^{\text{o}}}\text{C}$ is given by:
${{R}_{t}}={{R}_{0}}(1+\alpha t)$
Where, ${{R}_{0}}$ implies the resistance of the conductor at ${{0}^{\text{o}}}\text{C}$ and $\alpha$ is a constant called the ‘temperature coefficient of resistance’ of the conductor.

Complete step by step answer:
Resistance of the conductor at ${{50}^{\text{o}}}\text{C}$, ${{R}_{50}}=5\text{ }\Omega$, so,
$5={{R}_{0}}(1+\alpha (50))$….(1)
Resistance of the conductor at ${{100}^{\text{o}}}\text{C}$, ${{R}_{100}}=6\text{ }\Omega$, so,
$6={{R}_{0}}(1+\alpha (100))$…(2)
To calculate the value of $\alpha$, divide equation (2) by (1). Then we get,

& \dfrac{6}{5}=\dfrac{(1+100\alpha )}{(1+50\alpha )} \\\ &\Rightarrow 6(1+50\alpha )=5(1+100\alpha ) \\\ &\Rightarrow 6+300\alpha =5+500\alpha \\\ &\Rightarrow 1=200\alpha \\\ &\Rightarrow \alpha =\dfrac{1}{200}=0.005\text{ per}{{\text{ }}^{\text{o}}}\text{C} \\\ \end{aligned}$$ Put the value of $$\alpha $$ in either of the equation, (1) or (2), to get the value of resistance at $${{0}^{\text{o}}}\text{C}$$. Putting the value $$\alpha =0.005$$ in equation (1): $$\begin{aligned} & 5={{R}_{0}}(1+\alpha (50)) \\\ & \Rightarrow 5={{R}_{0}}(1+(0.005)(50)) \\\ &\Rightarrow 5={{R}_{0}}(1+0.25) \\\ & \Rightarrow 5=1.25{{R}_{0}} \\\ &\Rightarrow {{R}_{0}}=\dfrac{5}{1.25}=4 \\\ \end{aligned}$$ **Hence, the resistance of the conductor at $${{0}^{\text{o}}}\text{C}$$ is $$4$$ ohm.** **Additional Information:** The resistivity of a pure metal is directly proportional to its absolute temperature, provided the temperature coefficient of the metal is $${{(1/273)}^{\text{o}}}{{\text{C}}^{-1}}$$. The resistivity of alloys also increases with rise in temperature, but this increase is much smaller compared to pure metals. **Note:** With increase in temperature, the amplitude of vibrations of the positive ions of the metal increases and they further obstruct the path of free electrons. Hence, the resistance of the conductor increases, that is, resistivity of the material increases. In other words, with rise in temperature, the electric conductance of the conductor decreases.