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How will a car 10A circuit deal with double load?

The brake lamp circuit is generally fused at 10A on most passenger cars. Since incandescent light bulbs present inrush currents of about 10x their steady state currents, I am guessing these fuses are slow fuses. One such bulb will steadily draw 1.5 - 2A. I am looking to using that same circuit for lighting up a second similarly rated light bulb and got two contradicting thoughts:

- given the industry's cheapening of everything trend I would assume the wire gauge is just enough to withstand one bulb and not two.
- then again, modern cars also support trailers, so doubling the load on that circuit might be fine.

What are your thoughts ?
First, I doubt adding another bulb will have any undesirable affect.
But check the wire gauge to make sure. If in doubt, add a relay to make sure it’s ok.
I’ve added loads of extra lighting to my van including tow bar electrics.

The fuse protects the wire and nothing else.

If the fuse is rated at 10A then so is the cable - else they have 'broken the rules' on cable selection and/or fusing.

I very much doubt they'd risk litigation by not complying with the regs associated with low voltage DC wiring.
Well, I have taken some measurements and run some calculations on them today, which ultimately resulted in ~0.9 V drop on the wires. I guesstimated some 5m of wire run of my standard looking hatchback. According to this calculator the dimension that corresponds to that voltage drop over that distance is AWG 22.
This sounds about right to me; visually, the wires going into tail lights connector appear only a bit thicker than your average UTP cat 5E wires (AWG 24). Measuring these wires' outer diameter I got 1.4 mm (sheathing included).
So these are likely 22 AWG wires, which, if this page speaks truth, are only good up to 7A. Still good enough for my needs, but not what the recommendation on this page is.
Now, adding a second lamp into the circuit would further drop the voltage by another 0.9V, i.e. double the load = double the drop. I don't suppose this would matter all that much for some regular light bulbs, would it ? Oh, and, there will be an additional ~1.1V drop from my circuit, so... make that 2 * 0.9 + 1.1 = 2.9 V total drop. Hmm....
Are you assuming 12V?
When the engine is running, the alternator will be outputting >14.V.
Adding another bulb won’t make the slightest difference.

Good catch!
I have measured 11.8V at the tail lights connector without load, engine off. Battery was 12.2V. II need this working down to a battery voltage level of 10.5V which is about the time when car batteries get swapped. That would make probably 10.1V without load at tail lights connector from which a further 2.9V to be subtracted, leaving just 7.2V for the lamps.
For some reason I have not measured the actual current, but used the standard 1.75A assumption (21W @ 12V standard bulbs).
1.75A would be correct for 12V.
And 1.5A approx for 14V. The inrush you mentioned would be too quick for any wiring or switches to over heat. You have the fuse there for a reason.
Out of curiosity, what bulb is it? A brake light? Which is normally 21W or an indicator? Which is also 21W.

Previous measurements were from the right side brake lamp circuit. The add-on I am looking to build will be hooked up to the same brake lamp circuit. Today I have measured drawn amps for the same right side brake lamp: 1.56 A
Then masured DMM probes resistance at ~0.4 ohm which translates into ~ 0.62 voltage drop. Ad this to yesterday's 0.9V drop = 1.5V drop from yesterday's 11.8V (no load) -> lamp was running today on 1.56A @ 10.3V

Valid point as there are legal values for lighting in cars.....

Some data on minimum permitted lumens (Europe) would be nice. Also a graph for incandescent 21W bulbs vs voltage luminosity output...

As we all did years ago, for tow bar electrics.
Just do it. The bulbs will draw more current. The wire and fuse will be fine.
They will still glow below your minimum requirement just like a torch when batteries are low.