The resistance of a conductor is 5 ohm at 50*c and 6 ohm at 100*c.Its resistance at 0*is??THANK YOU!!

May 30, 2015

Well, try thinking about it this way: the resistance changed by only $1 \Omega$ over ${50}^{o} C$, which is a pretty large temperature range. So, I would say it's safe to assume the change in resistance with respect to temperature ($\frac{\Delta \Omega}{\Delta T}$) is pretty much linear.

$\frac{\Delta \Omega}{\Delta T} \approx \frac{1 \Omega}{{50}^{o} C}$

$\Delta \Omega = \frac{1 \Omega}{{100}^{o} C - {50}^{o} C} \cdot \left({0}^{o} C - {50}^{o} C\right) \approx - 1 \Omega$

${\Omega}_{{0}^{o} C} \approx 4 \Omega$

Jun 15, 2015

Its resistance at ${0}^{\circ} \text{C}$ is 4 ohm.

Explanation:

${R}_{T} = \left(1 + \alpha T\right) R$, where

${R}_{T} =$Resistance at any temperature,
$\alpha$=constant of material,
$R =$resistance at Zero degree Celsius.

At 50 degrees Celsius:

${R}_{50} = \left(1 + 50 \alpha\right) R$=$\text{5 ohm}$ $\text{ } \textcolor{b l u e}{\left(1\right)}$

At 100 degrees Celsius:

${R}_{100} = \left(1 + 100 \alpha\right) R = \text{6 ohm}$ $\text{ } \textcolor{b l u e}{\left(2\right)}$

At zero degrees Celsius:

${R}_{0} = \left(1 + 0\right) R$
${R}_{0} = R$ $\text{ } \textcolor{b l u e}{\left(3\right)}$

Determination R from equations $\textcolor{b l u e}{\left(1\right)}$ and $\textcolor{b l u e}{\left(2\right)}$** by

$\frac{\textcolor{b l u e}{\left(1\right)}}{\textcolor{b l u e}{\left(2\right)}} \implies \frac{1 + 50 \alpha}{1 + 100 \alpha} = \frac{5}{6}$

$6 + 300 \alpha = 5 + 500 \alpha \implies \alpha = \frac{1}{200}$

Use this value in equation $\textcolor{b l u e}{\left(1\right)}$

$\left(1 + \frac{1}{200} \cdot 50\right) \cdot R = 5 \implies \frac{5}{4} \cdot R = 5 \implies R = \text{4 ohm}$

According to equation $\textcolor{b l u e}{\left(3\right)}$, you have

${R}_{0} = R = \text{4 ohm}$

Therefore, its resistance at ${0}^{\circ} \text{C}$ is $\text{4 ohm}$.