# How do I calculate the power consumed by a lightbulb?

I'm studying a lightbulb and its variable resistance, given by the expression:

$R(T) = Ro[1 + α(T-T_0)]$, where $R_0$ is the resistance of the lamp at $T_0$.

In this case, $R$ is not given by Ohm's law ($V=Ri$). So, which expression can I use to calculate the power consumed by a lightbulb?

• $P = R\cdot i^2$
• $P = \frac{V^2}{R}$
• $P = V\cdot i$ (I don't think I can use this one)

I have measured the current and the e.p.d. at the bulb using instruments.

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All those expressions are equivalent and can be used, if you simultaneously measure, the voltage, current (and know the resistance from the temperature).

Why do you think ohm's law doesn't apply? Resistance is only dependent on temperature in your case (not current or voltage).

In general I would use P = IV, if Ohm's law actually doesn't apply.

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I think I was measuring the voltage incorrectly. Since at my calculations, the power was different on each case. Actually, I think we measured the epd including the protection resistance. Anyway, thanks for your reply! –  Fábio Perez Jun 29 '11 at 22:17

Ohm's law applies. Looks like the easiest form to use would be ${V^2}/{R}$.

The part which you are missing, and which makes it impossible to solve this even in the steady state, is the equation for heat dissipation versus temperature in the lamp.

To solve the steady state problem, you need thermal resistance (or conductivity) to ambient $T_0$ (as well as knowing ambient $T_0$). You need all this to come up with a steady state T, which allows you to solve for R. The real world solution involves radiant, convective and conductive losses, so will appear to be a non-linear equation. Once you have that equation in hand, you will almost certainly have to iterate to close in on a solution: guess at R; get power; solve for T; use that T to calculate a new R; repeat; repeat; repeat...

To solve the transient problem, you additionally need the specific heat (heat storage capacity per $T_{delta}$) of all elements in the thermal path.

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