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I need to replace the taillight/brake light on my car, and it's a bulb with two filaments (known as 12V P21/5W - see pic)

My car has two taillights on each side. A standard 5W bulb for the top half, and the bottom one also doubles as a brake light when you use the brakes. So if you have the lights on, it lights up like the top half of the taillight (a normal 5W bulb) but when you press the brakes, it also lights up the other filament (maybe both at the same time? I'm not sure)

enter image description here

My question is, since it's connected like a normal light bulb, how does it know which filament to light up? Does it depend on how much power is provided?

cantsay
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5 Answers5

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A typical 1157 taillight bulb looks like this:

enter image description here

The filaments have a common connected to the shell and the other end of each goes to each the two terminals on the bottom of the bulb. The bayonet pins are offset so the bulb can only be inserted one way.

In the case of the 1157 bulb, the elements are 26.8W/8.3W, and have about a 10:1 difference in lumens (and about 1/4 the lifetime for the brighter brake light filament).

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This kind of light bulb has are more than two contacts.

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This kind of light bulb has three contact and is keyed to only go in one way. The first contact is the silver portion of the base of the bulb with the knob. This is the ground and it also acts as the key. The knob on the other side is higher up and that only lets the bulb be installed in one orientation. The other two contacts are on the black underside.

vini_i
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I was looking for the answer, but there is no full information in other answers.

Look at my solution.

Sorry for watermarks but I haven't found clear picture.

enter image description here

0

As @ndim said, this light is probably not connected like a normal light bulb.

However, if it was (which is possible), it could have a power FET or some other sort of switch. That way, if enough power was put through, it would switch on and allow current through the second filament as well as the first.