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In physics we were learning about how in series the current is the same through each resistor. So that the lightbulbs will all have the same brightness but dimmer than if there was just one bulb on there. We were explained the current being the same by Ohm's law: $V = IR$. As the resistance increases from the wire to the resistor the voltage also increases, cancelling each other out in the equation $I = \frac{V}{R}$. Leaving it as being the same.

But my question is: doesn't this only apply for Ohmic resistors, and lightbulbs aren't ohmic resistors? Would the first light bulb be brighter than the ones after it?

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  • $\begingroup$ Old fashioned, incandescent lightbulbs are Ohmic (though they have a temperature dependent resistance). Fluorescent (including compact fluorescent) and LED bulbs are another matter. $\endgroup$ – dmckee May 8 '16 at 3:38
  • $\begingroup$ At AC household frequencies, incandescent bulb filaments reach a stable temperature and a stable operating resistance. For some plots of voltage vs current at low frequencies so that the temperature is not stable see physics.stackexchange.com/questions/214897/… $\endgroup$ – Bill N Apr 18 '18 at 13:31
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Light bulbs, or any loads, in series will all have the same current. This is unrelated to Ohm's Law - it's Kirchhoff's Current Law and it applies if the loads are ohmic or not. Assuming your source voltage stays the same, adding bulbs in series will increase the total resistance which will decrease the total current and make all the bulbs dimmer. The order of the bulbs is not significant. If one of the bulbs in series has a higher resistance it will be brighter.

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You are correct that light bulbs are non-ohmic (they don't obey Ohm's Law). But that makes no difference. The same current flows through each, even if they have completely different resistances.

Electrical current (charge per second) is like the flow of a river. If there are no leaks, and no tributaries joining the river, then the volume of water per minute passing a point upstream is the same as the volume per minute passing a point downstream. The water in the river cannot disappear, neither can the current leak out of the wires. Every second, what goes in must come out. So the same electric current flows through all components which are in series.

If the bulbs are identical and the same current flows through each, then they will have equal brightness. It doesn't depend on Ohm's Law. All it needs is for the resistances to be equal and the current to be the same.

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I can't understand at all everything after "As the resistance increases ...". Is that really how it was explained?

Nonetheless there's an interesting point here. The analysis is not exactly the same as for an ohmic resistor, but not for the reason you suggest.

The thing to consider is that the resistance of the light bulb depends on the temperature of the bulb, and we don't have complete information about that. One thing is certain: the current through the two bulbs will be exactly the same. But the light bulb with the higher resistance will be brighter. If the light bulbs are absolutely identical, then they will heat up identically, and reach a common final temperature, and the bulbs will indeed have the same brightness.

Real light bulbs will not be identical (although they are usually close) so one will be at a different temperature, and the brightnesses will be different.

But there's no reason to think that it will be the "first" bulb. In fact, how do you determine which is "first"? The only reason one would be brighter than the other is because there are inevitable small variations in the manufacturing process.

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in series the current is the same through each resistor.

Not just the same, the current through each is identical.

So that the lightbulbs will all have the same brightness but dimmer than if there was just one bulb on there.

Well of course, series connected resistances add and so, the total resistance of two series connected bulbs is greater than the resistance of either individual bulb.

We were explained the current being the same by Ohm's law: V=IR.

Have you properly identified the $V$ and $R$ in this equation? Each bulb has a voltage across and a current through. We know the current through is identical since the bulbs are series connected.

However, the voltage across each bulb can be, in principle, different since, in general, the bulbs are not identical.

Assume you have two bulbs that are ohmic and have resistances $R_1$ and $R_2$ respectively.

Since the two bulbs are series connected, it follows that the series current $I$ is

$$I = \frac{V_1}{R_1} = \frac{V_2}{R_2}$$

Then, by KVL, the total applied voltage is

$$V = V_1 + V_2 = IR_1 + IR_2 = I(R_1 + R_2)$$

and thus,

$$I = \frac{V}{R_1 + R_2}$$

Now, the brightness of either bulb is proportional to the current $I$ through and it is clear that $I$ is less when the two bulbs are series connected than either bulb by itself.

Thus, we conclude, the two bulbs in series will be dimmer than if there were "just one bulb on there".

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