I thought that if the voltage source is 120V and the lightbulb is 40W then the current would be 1/3 of an ampere meaning that the resistance of the lightbulb would be 360 ohms. But when I checked it with my multimeter, it was only 26 ohms. The multimeter is not wrong. If I check a 1K resistor with the multimeter it reads 1K.

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    \$\begingroup\$ @WesleyLee Actually they are Ohmic (the conductor is a tungsten filament, which is an ohmic material). The problem is that resistivity is a function of temperature, so Ohm's law is still valid, but the bulb must reach thermodynamic equilibrium first. \$\endgroup\$ Jun 19, 2021 at 5:59
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    \$\begingroup\$ @WesleyLee In other words, Ohm's law is not independent of temperature. The proportionality between V and I is guaranteed (for an Ohmic material) as long as the dissipated power doesn't change the thermal equilibrium appreciably, otherwise any material would be non-ohmic, since its resistivity will change on a sufficiently wide temperature range . \$\endgroup\$ Jun 19, 2021 at 6:07
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    \$\begingroup\$ Non-ohmic materials are those whose resistivity changes appreciably with voltage (or current), even when held at constant temperature. \$\endgroup\$ Jun 19, 2021 at 6:09
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    \$\begingroup\$ @LorenzoDonati--Codidact.com -- thank you very much for your explanation! Since I cannot edit my comment I will delete it. Dont want to keep wrong info on the site :) \$\endgroup\$
    – Wesley Lee
    Jun 19, 2021 at 7:58
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    \$\begingroup\$ this is why ham operators use in-series lightbulbs when debugging; it's electrically near-invisible at small currents (bulb appears off) and if there's an unexpected big current, the bulb limits the current and glows a warning. \$\endgroup\$
    – dandavis
    Jun 19, 2021 at 8:55

5 Answers 5


Filaments heat up to produce light and the tungsten they're made of will change resistance as this happens. The temperature change is let's say 3000K, the temperature coefficient of resistance for tungsten is 0.0045/K, so the resistance when the bulb is on will be about 13.5x what it is when it is cold. Plugging in your measured 26 ohms gives 351 ohms when hot.

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    \$\begingroup\$ I would love to see the graph of the current through such a bulb, starting from cold. The bulbs take a clearly visible time (probably 0.1 to 0.2 seconds) to ramp up from zero to full brightness. There must be one hell of a starting current. \$\endgroup\$
    – paki eng
    Jun 21, 2021 at 8:17
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    \$\begingroup\$ @AbdullahBaig You bet there is some overshoot. And that's why light bulbs burn out when they are switched on. \$\endgroup\$ Jun 21, 2021 at 8:57
  • \$\begingroup\$ I wonder why the filament would burn out. The burn out would be most likely due to melting and the heat dissipation is almost 100% through radiation (which is linked to temperature without any delay). Hotspots due to asymmetries? \$\endgroup\$
    – paki eng
    Jun 21, 2021 at 11:29
  • \$\begingroup\$ @AbdullahBaig probably that, I remember that if the bulb got hit or shaken it would burn out within a couple days because the filament would loop over itself \$\endgroup\$ Jun 21, 2021 at 14:25

The resistance of a lightbulb is not constant. Becomes the lightbulb resistance is not constant but will change with the applied voltage. See the graph of resistance of a \$100W/230V\$ light bulb in the function of supply voltages (and temperature of a filament).

enter image description here

The resistance for \$230V\$ is equal \$530Ω\$ so current is equal \$0.434A\$ But for \$120V\$ the current will not be equal to:

\$I = \frac{120V}{530Ω} = 0.226A\$

As we can read from the graph the current will be equal to:

\$I = \frac{120V}{380Ω} = 0.315A\$

Also, notice that the cold resistance is equal to \$40Ω\$. And the hot resistance at \$230V\$ is \$530Ω\$. Thus, the ratio between "hot"/"cold" resistance is \$\frac{530Ω}{40Ω} \approx 13.2\$

And in your case, we have a similar situation, the resistance ratio is 360Ω/26Ω is also around 13.8

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    \$\begingroup\$ Does this mean that the light bulb will consume much more than 40W for a few instants, until it gets hot? How many milliseconds does that usually take? \$\endgroup\$ Jun 19, 2021 at 17:52
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    \$\begingroup\$ @RazvanSocol This is exactly what will happen. As for the time for heat up, it will depend on time bulb time contact (around 50ms). And what is the current value of an AC voltage? Is it at the peak value, zero crossings of between? \$\endgroup\$
    – G36
    Jun 19, 2021 at 18:24
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    \$\begingroup\$ @RazvanSocol It's why incandescent lightbulbs often break at startup, especially an outdoor bulb in a cold winter. In fact, you can greatly increase the bulb lifetime by adding a soft-start circuit to limit the startup current, a simple thermistor will do. It's something you can buy, the Bulb-Miser, it's just a small disc that you insert between the bulb base and the socket, a cheap, simple and effective device. \$\endgroup\$ Jun 20, 2021 at 21:32
  • \$\begingroup\$ OK on all of the above. What if two such light bulbs are in series. If one has a bit more resistance, it will have a higher voltage, dissipate more power, and rob the other bulb of voltage and power. A positive feedback situation exists. I would expect on bulb to be much brighter than the other. And if a surge of power could be applied only to the dimmer bulb, it could permanently starve its competitor, resulting in a primitive memory device. Why do I not see this? K6YVL \$\endgroup\$ Jun 25, 2021 at 15:55
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    \$\begingroup\$ @G36 Your are correct about the lower power light bulb running brighter. However, I was looking at two equal light bulbs. If one has a little higher resistance, it starves the other of power and causes it to run cooler. Why don't I see this with light bulbs in series? It is probably time for some experiments to characterize the resistance curve and then run some simulations.. It this really is a two-state system, another way to change the state might be to reflect light back onto the filament of one light bulb, or even give it a burst of heat with a laser. \$\endgroup\$ Jun 27, 2021 at 5:09

The cold resistance of a incandescent light bulb is considerably lower than the running resistance it has during use when hot.

A rough estimate is that hot resistance is about 10x larger than the cold resistance. It also means the switches and wires see roughly 10x larger current surge when the light is turned on.

So a 40W bulb at 120V would be 360 ohms, and you measure 26 ohms when cold. That is approximately in the expected range, the hot to cold ratio is 13.8x in your case.

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    \$\begingroup\$ The low resistance when cold is part of why incandescent lights are so much more likely to burn out when switched on than when already operating, if not using a dimmer switch for gradual-on. A thin spot will have higher resistance and heat up more, getting even more resistance and thus a larger share of the total power being dissipated, before the surrounding parts also get hot and drop more of the voltage. If that runaway effect lets one portion get too hot during the high-current startup surge, you can get a fuse-like effect. \$\endgroup\$ Jun 19, 2021 at 11:56
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    \$\begingroup\$ @PeterCordes: Interestingly, DJL projector bulbs (and likely some others) have a filament in the back which is in series with the main filament, but doesn't get nearly as hot (it's not positioned in front of the reflector, so none of its light is actually useful). My suspicion is that on startup, the secondary filament's resistance is probably somewhat close to the primary filament, thus limiting inrush current, even though after that the secondary filament represents a pure waste of energy. \$\endgroup\$
    – supercat
    Jun 20, 2021 at 17:41

As other answers has already explained, as the filament heats up it will increase its resistance.

There is one more parameter that you possibly would take inte account. As the light bulb is connected to AC a number of other effects may come inte play. I list them in no particular order:

  • 120V AC: how is this 120V actually defined? Ideally it should perhaps be RMS valde, meaning that the maximum voltage will be higher. And an AC signal passes by zero voltage each period.
  • How far from a perfect resistor will the filament actually be: a coiled filament will have at least a small amount of reactance. In effect there will be a lag between voltage and current through the filament. In power electrics this can be described by the power factor, the ratio between working power (that is the light and heat) and the appearant power (that mostly creates heat in other places). Admittedly for a typical bulb it is very close to one.
  • A small tidbit of info, a typical efficiency would be around 12 lm/W when run at designed voltage. At lower voltage, the bulb may create next to no light of interest to human eyes, only heat.

The unstated assumption in your question is that Ohm's law (\$V=IR\$) is applicable to any device you can put an ohmmeter on.

This is not true. Here are some examples of devices where it does not hold:

  • switches
  • batteries
  • diodes
  • transistors
  • inductors
  • capacitors
  • gas arc lights

All of these things have some relationship between current and voltage. For example, an ideal capacitor:

$$ i(t) = C{\mathrm d v(t) \over \mathrm d t} $$

This equation isn't a law named after anyone: it's simply a definition of capacitance. When we have some device which has some behavior which can be sufficiently modeled by this equation, we call it a capacitor with capacitance \$C\$.

Likewise, Ohm's law is a definition of resistance:

$$ V = I R $$

This equation is simpler, and it has a name, but regardless it's only applicable to resistors.

Other answers have said a light bulb's resistance increases with temperature. While this is true, I don't think it's a good enough explanation, because it still suggests maybe all things are somehow subject to Ohm's law, when quite clearly they are not. What good is a law that says current and voltage are related by some ratio called "resistance" when resistance can change?

There's a bit of a subtle trick here: while "resistance" and consequently Ohm's law only applies to resistors, there's a second use for the term in creating simplified models for devices or circuits. Consider the Shockley diode equation for example, which describes how current and voltage are related for an ideal diode:

$$ {\displaystyle I=I_{\mathrm {S} }\left(e^{\frac {V_{\text{D}}}{nV_{\text{T}}}}-1\right)} $$

This is a rather complicated equation. Many of the terms are parameters that will be particular to a given diode (and probably vary with temperature and frequency), but even after simplifying those you're left with an exponential relationship between current (\$I\$) and voltage (\$V_\text D\$). If you plot voltage and current you get something like this:

enter image description here
Thingmaker, CC BY-SA 4.0, via Wikimedia Commons

While it's essential to understand the exponential relationship exists in general, you may only be interested with the behavior of the diode over a very narrow range, say from 0.78 to 0.8 volts. On such a small scale, it might be "good enough" to call this exponential curve a linear one. Then you can take the slope of that curve and call it "resistance", and now you can substitute the complex Shockley model for a simpler linear one as long as you stay within that assumption of 0.78 to 0.8 volts.

In other words, it's not that a diode obeys Ohm's law in any way, but that you've elected to make a simpler model of the device which is sufficiently accurate within a range of assumptions. If you know the diode is operating within 0.78 to 0.8 volts, and the "resistance" is 13 ohms, then you can easily calculate that if the current increases by 0.1 mA then the voltage will increase by \$0.1\:\mathrm{mA} \cdot 13\:\Omega = 1.3\:\mathrm{mV}\$.

When someone says "the resistance of a light bulb increases with temperature", the situation is similar. Light bulbs have a current-voltage relationship which depends on temperature to a large extent. Modeling that temperature-dependent behavior may be unnecessarily complex for the task at hand, and so engineers, not wanting to make things harder than they need to be, have determined that assuming the temperature of the filament is reasonably constant, the current-voltage relationship of a light-bulb can be approximated by a simple linear equation. It's not that light bulbs obey any "Ohm's law": it's that someone has taken a complicated model of a light bulb and made a simpler one.

  • \$\begingroup\$ "This equation is simpler, and it has a name, but regardless it's only applicable to resistors." Actually, it is applicable to many problems in physics, such as magnetic flux, gas flowing through a hole, and immigrants crossing the U.S. border. \$\endgroup\$ Jun 27, 2021 at 5:14

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