Question

If resistance of wire at ${50}^{\circ }$C is 5 R $\Omega$ and ${100}^{\circ }$C is 6R $\Omega$, find resistance of $0^{\circ }$C.

Answer 1

We know that,

$R_t=R_0(1+\alpha t)$

Where $R_1$ is the resistance of the wire at $t^{0}C$

$R_0$ is the resistance of the wire at $0^{0}C$

and $\alpha$ is the temperature coifficient of resistance.

$\Rightarrow R_{50}=R_0(1+50\alpha )-----\left(i\right)$

$R_{100}=R_0(1+100\alpha )------\left(ii\right)$

From $\left(i\right)$, $R_{50}-R_0=50\alpha R_0---\left(iii\right)$

From $\left(ii\right)$, $R_{100}-R_0=100\alpha R_0----\left(iv\right)$

Dividing $\left(iii\right)$ by $\left(iv\right)$, we get

$\frac{R_{50}-R_0}{R_{100}-R_0}=\frac{1}{2}$

Here $R_{50}=5\Omega$ and $R_{100}=6\Omega$

$\therefore \frac{5-R_0}{6-R_0}=\frac{1}{2}$

or $6-R_0=10-2R_0$ or, $R_0=4\Omega$