91Ó°ÊÓ

According to Ohm’s law the current flowing through conductor is directly proportional to the voltage across the two points.

$V=IR$

Where, $I$is current in ampere $\left(A\right)$, R is resistance in ohms $\left(\mathrm{Î\copyright }\right)$ and V is the potential difference volt role="math" localid="1664174969095" $\left(V\right)$.

The ratio of V to I for a particular conductor is called its resistance R :

$R=\frac{V}{I}$or $\frac{\mathrm{ÒÏ}L}{A}$

Where, $\mathrm{ÒÏ}$is resistivity $\mathrm{Î\copyright }·\mathrm{m}$, $L$is length in m and$A$ is area in $m^2$.

Also,$R=R_0\left[1+\mathrm{Î\pm }\left(T-T_0\right)\right]$

The power$\left(P\right)$ is the product of potential difference$\mathbf{\left(}\mathbf{V}\mathbf{\right)}$ and the current$\mathit{(}\mathit{}\mathit{I}\mathit{}\mathit{)}$ .

$P=VI\mathrm{or}{I}^{2}R\mathrm{or}\frac{V^2}{R}$

Given that, at temperature$80.0°C$ the power is 480 W , the reference temperature$T_0$ is$20.0°C$ and the coefficient of tungsten$\mathrm{Î\pm }$ is$0.0045{\left(C°\right)}^{-1}$ .

The resistance at temperature$80.0°C$is

localid="1664175441780" $$\begin{gathered} R=R_0\left[1+\mathrm{Î\pm }\left(T-T_0\right)\right] \\ =R_0\left[1+0.0045\left(80-20\right)\right] \\ =1.27R_0 \end{gathered}$$

Now, the potential difference is

$$\begin{gathered} P=\frac{V^2}{R} \\ 480=\frac{V^2}{1.27R_0} \\ V=\sqrt{480[1.27R_0]} \\ V=24.6\sqrt{R_0} \end{gathered}$$

Hence, potential difference at temperature$80.0°C$ is $24.6\sqrt{R_0}$.

The resistance at temperature is

$$\begin{gathered} R=R_0\left[1+\mathrm{Î\pm }\left(T-T_0\right)\right] \\ =R_0\left[1+0.0045\left(150-20\right)\right] \\ =1.58R_0 \end{gathered}$$

Now, the power at temperature is

$$\begin{gathered} P=\frac{V^2}{R} \\ =\frac{{\left(24.6\sqrt{R_0}\right)}^2}{1.58R_0} \\ =\frac{605.2R_0}{1.58R_0} \\ =383\mathrm{W} \end{gathered}$$

Hence the power at temperature$150.0°C$ is 383 W .