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Increasing resistance with temperature.

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When a bulb, connected across potential $V$ of initial resistance $R_o$ is switched on its resistance increases due to heat production. Assuming initial temperatures of the bulb is $0^{\circ}C$, and temperature coefficient of the bulb is $\alpha$. After what time its temperature will become $T$. The heat capacity of the bulb is $c$.

I only know that $R=R_o\cdot e^{\alpha\cdot (T_2-T_1)}$ how do I apply it here, please help?

asked Apr 3 in Physics Problems by n3 (448 points)
edited Apr 6 by sammy gerbil
This can be solved using differential equations.

The power delivered to the filament of the bulb is $P=\frac{V^2}{R}$. Assuming there are no heat losses, the power delivered to the bulb is related to the temperature increase of the filament by $P=c\frac{dT}{dt}$ where $c$ is the heat capacity of the bulb and $T$ is the temperature increase.

Usually the temperature coefficient assumes linearity : $R=R_0(1+\alpha T)$. For small deviations $T$ of temperature from the initial value, such that $\alpha T \ll 1$, this is the same as the exponential law : $$R=R_0e^{\alpha T} \approx R_0(1+\alpha T+\frac12 \alpha^2 T^2+...)$$

1 Answer

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The power delivered to the bulb is $P=V^2/R$. Assuming there are no heat losses, the power delivered to the bulb is related to the temperature increase of the filament by $P=c\frac{dT}{dt}$ where $c$ is the heat capacity of the bulb and $T$ is the temperature increase. Following your suggestion I shall model the variation of resistance with temperature as $R=R_0e^{\alpha T}$ where temperature $T=0$ at time $t=0$. Then $$\frac{dt}{dT}=\frac{cR_0}{V^2}e^{\alpha T}$$ which can be integrated to get $$t=\frac{cR_0}{\alpha V^2}(e^{\alpha T}-1)$$ For small values of $T$ such that $\alpha T \ll 1$ we get the approximation $$t\approx \frac{cR_0}{V^2}T$$

answered Apr 15 by sammy gerbil (27,948 points)
edited Apr 15 by sammy gerbil
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