Chapter 18, Problem 19P

To determine

The temperature of the filament when hot assuming an average temperature coefficient of resistivity $\alpha =0.0060{\left(C°\right)}^{-1}$ .

Answer to Problem 19P

  $1797°C$

Explanation of Solution

Given info:

A $100W$ lightbulb has a resistance of about $12\Omega$ when cold $\left(20°C\right)$ and $140\Omega$ on (hot).

An average temperature coefficient of resistivity $\alpha =0.0060{\left(C°\right)}^{-1}$

Formula used:

The resistivity of metal increases linearly with the temperature. Expression for this: ${\rho }_T={\rho }_0\left[1+\alpha \left(T-T_0\right)\right]$

Where,

  ${\rho }_0$ is the resistivity at reference temp $T_0$ ,

  ${\rho }_T$ is the resistivity at reference temp $T$ ,

  $\alpha$ is the true coefficient of resistivity constant.

Calculation:

Formula for temperature-dependent resistivity, ${\rho }_T={\rho }_0\left[1+\alpha \left(T-T_0\right)\right]$

Now, solve for $T$

  $T=T_0+\frac{\frac{{\rho }_T}{{\rho }_0}-1}{\alpha }$

The length and cross-sectional area are constant. Because of change in resistivity the resistance will be changed; $R=\frac{\rho L}{A}$

It means that the ratio of the hot as well as cold resistances are qual to the ratio of the hot and cold resistivities. So, $\frac{R_{hot}}{R_0}=\frac{{\rho }_T}{{\rho }_0}$

Substitute the resistances for the resistivities,

  $T=T_0+\frac{\frac{R_{hot}}{R_0}-1}{\alpha }$

Substitute the given resistances and the temperature coefficient $\alpha$ . So,

  $T=20°C+\frac{\frac{140\Omega }{12\Omega }-1}{0.0060}=1797°C$

Conclusion: The required temperature is $1797°C$ .