## Related questions with answers

A 100-W lightbulb has a resistance of about $12 \Omega$ when cold $(20^{\circ} \mathrm{C})$ and $140 \Omega$ when on (hot). Estimate the temperature of the filament when hot assuming an average temperature coefficient of resistivity $\alpha=0.0060\left(\mathrm{C}^{\circ}\right)^{-1}$.

Solution

VerifiedAccording to the equation (18-3):

$\rho_T=\rho_0[1+\alpha(T-T_0)].$

Now multiply this equation with $L/A$ (see (18-3)) to obtain temperature dependence of resistance instead:

$R_T=R_0[1+\alpha(T-T_0)].$

Solve for $T$:

$\frac{R}{R_0}-1=\alpha(T-T_0),$

$T=T_0+\frac{1}{\alpha}\left(\frac{R}{R_0}-1\right).$

Substitute the givens:

$T=20\text{\textdegree} C+\frac{1}{0.0068(\text{\textdegree} C)^{-1}}\left( \frac{140\Omega}{12\Omega}-1 \right),$

$T=1798\text{\textdegree} C \approx 1800\text{\textdegree} C.$

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