I am learning about voltage dividers.

In simple words, a voltage divider is a specific circuit setup where we connect multiple passive elements in series, in order to divide the main voltage into small voltage drops.

The voltage drop for each element is proportional to its resistance.

If I have a lamp that needs 6 V (and not more), but I have a 12 V supply, how is the voltage divider really created?

I could replace the lamp with a resistance of let’s say 3 ohms, and then say that I need another resistance connected with the lamp.

But this question arises:

  • Can a lamp (I intend the common Joule effect based lamp, not any other type of lamp) actually be “just” considered a resistance? The resistance will change with temperature and temperature will definitely change.

I guess the answer is no but I don’t know the exact reason. Also, being that the case... how would the circuit be in this case? I guess something like this makes most sense:

enter image description here

But again that changes the circuit and I don’t know if that setup will actually deliver 6 V to the lamp. In case this setup is wrong, how would I achieve the expected solution? To deliver only 6 V to that element. What are the real magnitudes that need to be known in order to make this happen in reality?


Problem with using a voltage divider is while you limit the voltage to the lamp to 6 volts, you also limit the available current from the 12 volt source to operate your lamp. The other 3 Ohm resistor (not the one in parallel) limits the maximum available current to the lamp to 12/3 = 4 amperes. And once your lamp starts drawing current the voltage across the lamp will drops. How much current does your lamp require (watts/voltage for resistive lamp filament)?

Hope this helps.

  • $\begingroup$ Thanks for your answer. This leads me to a bigger confusion. Not because your answer is not amazing, but because I did not know that a lamp needs a specific current to work. I thought the restricting parameter for the lamp to function is to give it that exact voltage. I will think and research and then see what the original question turns into. Will comment here again. $\endgroup$ Dec 15 '19 at 16:47
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    $\begingroup$ @AlvaroFranz The lamp should have a current or wattage rating in addition to a voltage rating. If its a resistive filament you can divide the wattage rating to get the current it is intended to draw. Depending on how much current it is intended to draw, you may be able to make adjustments to the voltage divider to ensure you have both the right voltage and current under load. $\endgroup$
    – Bob D
    Dec 15 '19 at 16:58
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    $\begingroup$ @AlvaroFranz "I thought the restricting parameter for the lamp to function is to give it that exact voltage". That is true except the voltage rating assumes there is no other resistance between the voltage source and the lamp, i.e., the voltage source is connected directly across the lamp and only the internal resistance of the source (which is normally low) would limit the current. $\endgroup$
    – Bob D
    Dec 15 '19 at 21:38

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