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Question

A 100-W lightbulb has a resistance of about 12Ω12 \Omega when cold (20C)(20^{\circ} \mathrm{C}) and 140Ω140 \Omega when on (hot). Estimate the temperature of the filament when hot assuming an average temperature coefficient of resistivity α=0.0060(C)1\alpha=0.0060\left(\mathrm{C}^{\circ}\right)^{-1}.

Solution

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Answered 2 years ago
Answered 2 years ago
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According to the equation (18-3):

ρT=ρ0[1+α(TT0)].\rho_T=\rho_0[1+\alpha(T-T_0)].

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

RT=R0[1+α(TT0)].R_T=R_0[1+\alpha(T-T_0)].

Solve for TT:

RR01=α(TT0),\frac{R}{R_0}-1=\alpha(T-T_0),

T=T0+1α(RR01).T=T_0+\frac{1}{\alpha}\left(\frac{R}{R_0}-1\right).

Substitute the givens:

T=20°C+10.0068(°C)1(140Ω12Ω1),T=20\text{\textdegree} C+\frac{1}{0.0068(\text{\textdegree} C)^{-1}}\left( \frac{140\Omega}{12\Omega}-1 \right),

T=1798°C1800°C.T=1798\text{\textdegree} C \approx 1800\text{\textdegree} C.

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