mercedes city light resistor watts?

I have a E430 Mercedes and the city lights inside the headlight assembly on the right side has a blown resistor.  I openeded up the other headlight to see what resistor to use and found a

Light Blue resistor

Color band:Red Yellow Black Silver Brown=2.4ohm

How would i find out what watts is the resistor?  The bulb is 12v 6w.

I tried to calculate it myself but i must be wrong.  Accoring to the ohms law regarding

Watts=Volts squaRED / ohms : 14.00volts squaRED / 2.4ohms=11.6watts Can't be right? 

Light blue means it is either flame-proof and fusible or metal film. I believe a glossy blue is metal film and a flat is flame-proof and fusible. If it has blown, check for a short before replacing.This information is provided only as a reference.

All circuits should be verified with a digital multi-meter prior to making any connections.

answer is 1/2 Watt.

It should probably be 1 Watt. 1/2W is a bit low - especially if charging above 14V. (The bulb could be 24R, hence 26.4R total = 8W @ 14.5V which means 0.7W from the resistor.    But if 0.5A bulb current assumed at 13.8V, then resistor wattage is 0.49W - too close to 1W?) PS: R = ohm.

How exactly did you calculate that oldspark, i never know i may need to know the correct way to find this out in the future.

I just cant win.... Either I write assumptions and derivations, or I skip them to minimise criticism. But I prefer the latter. If people (or rather the OP) want to know, they merely ask. (If not educated enough to understand nor understand their "right to request" more info.... - that's not MY problem. I sound like a politician by replying "I'm glad you asked me that question...".   But (1) I'm genuine - especially since you are the OP, & (2) I'm not sure "how glad" (LOL) - will I remember how, or find errors, or reduce my "new" answer to under half the verbiage etc? (Ok, not the latter because I gave the Executive answer. Phew!) First - the assumptions [and extra (trivial?) info italicised in square brackets]:    Assume our system varies from 12V to 14.4V - the "nominal" system voltage and normal "maximum" charging voltage of a 12V vehicle. [For design purposes, I use a range of 8V to 16V. Under 12V doesn't worry us in this case, but we could use 16V.]    A 12V system is nominally "rated at" 13.8V - the float voltage of a 12V lead acid battery (at 25 degrees C). [Ideally: Charge at 14.4V till full, then drop to float voltage of 13.8V.] 12V components are normally "rated" at 13.8V [so is your bulb 6W @ 12V or 6W @ 13.8V?] Use the worst-case scenario, but be practical [ie, be aware that real component ratings vary (eg, 6W = 5W to 7W), and do a final sanity check in case "reality" might be a lot cheaper]. Ohms Law & power formulae: V=IR & P=VI [hence P=IIR = VV/R this R=VV/P (where VV = VxV = "V-squared" = =V² or V**2 or V^2 etc; same for II). And I'll use "R" instead of O. City lights might be dimmed "normal" bulbs or an under-stressed bulb, hence the resistor is in SERIES with the bulb. [Hence it is not 14V across the resistor! P=VV/R = 14x14/2.4 = 81.666... = 82W. And your lamp failed when the resistor blew.] Now, the calcs.... finally (not) The bulb could be 24R - ie, 6W@12V => 144V/6W = 24R. Bulb plus series resistor = 24+2.4 = 26.4R. From this we get our first sanity check (aka approximation)... Because the resistor is 1/10th the bulb resistance, it will dissipate 1/10th the bulb wattage = ie, 0.6W. [2.4R/24R = 0.1; 0.1 x 6W = 0.6W] ... and the extra bit & more calcs.... Normally I would leave it there. With topinstaller200's reply of 0.5W I feel sane enough, but would call it 1W. [Resistors being eg 1/4W, 1/2W, 1W, 2W, 5W, 10W, 20W.] [Although the added resistor reduces the lamp current hence overall current hence overall power, this is only by about 10% (being 2.4R compared to 24R) so that means 10% less than 0.6W = 0.54W. Still > 0.5W => a 1W resistor. Of course it's actually 1/11th from 10+1=11 (2.4 in 26.4 is one-eleventh; ie, 1 in (1+10) = 1 in 11; or 2.4+24 = 2.4(1+10) etc)] But being pedantic, I try other values and methods too.... Let's try 14.5V. P=VV/R = 14.5x14.5/26.4 = 210.25/26.4 = 7.96W = 8W. The resistor dissipates 2.4/26.4 of this = 8W/11 = 0.72W Hence my "hence 26.4R total = 8W @ 14.5V which means 0.7W from the resistor." Or from the 1:10 approximation, 8W/10 = 0.8W - still above 0.5W and not "borderline" with 1W. If the bulb is a fixed resistance of 24R being 6W@12V [it isn't, but its resistance increases with increasing voltage - hence the "thermal self limiting" and why bulbs (and many other things) don't blow considering their lower resistance when cold], that means 0.5A @ 12V [I=P/V = 6W/12V = (1/2)A.] As to my "But if 0.5A bulb current assumed at 13.8V, then resistor wattage is 0.49W - too close to 1W?), I was being conservative and assuming 0.5A @ 13.8V - that 13.8V being the lowest "maximum" voltage that any vehicle should have. (Certainly voltages above a full battery voltage up to 12.8V are assumed!) But that is flawed and a bit complex - I think I did something like "if the 12V/6W means 0.5A and that is the current @ 13.8V".... then dropping current by 1/10th means 0.45A => .45x.45x2.4 = 0.486W - 0.49W. My too close to 1W? should be "too close to 1/2W?. So there is my d'oh typo. The above "0.5A@13.8V" was also my way of allowing for higher voltages (13.8 + 10%), but a better method is to assume a 6W bulb @ 13.8V => 31.72 = ~32R. 14.4V across (32+2.4)R => 0.42A, hence .42x.42x2.4 = .43W from the resistor. Hence a 1/2W resistor. Thereth the lesson ends. But I'll continue anyhow.... I could have said at the start regarding...

fiat1980spyder wrote: Watts=Volts squaRED / ohms : 14.00volts squaRED / 2.4ohms=11.6watts Can't be right?

...isn't right. [4.00volts squaRED / 2.4ohms is NOT 14x2/2.4 = 28/2.4 = 11.6. It is 14x14/2.4 = 196/2.4 = 81.666 = 82W.

That's err#1.

Err#2 is that the resistor is not parallel to the bulb (hence receiving 14V), but is in series. Think logical - how does a resistor (restriction) alongside a component (parallel to a water sink) reduce the current flow through that component (reduce the water flow - the restriction or tap has to be in line!)] But what's the fun in that? Besides, I want others not to make the same mistakes I have made time & time again. I mean, the error you made. (Sorry, now I am guilty of transference too! But Geez, how may times? D'oh!!) And this way I show how such calcs are done. And that for most experienced people (does the name "topinstaller200" suggest something?"), that is the "correct" answer. IE - 0.43W, and that means a 0.5W resistor. Even though .43 is close to .5, that is at 14.4V which is the "theoretical real maximum" that a vehicles should see. And in practice - especially older cars - many don't have 14.4V with even light loads (no pun). In summary, the 1/2W answer assumes the normal real world ratings that "12V/6w" means 6W @ 13.8V. And that is probably correct.     And why chose 1Watt? But I'd be tempted to use a 1W resistor unless there was a reason not to - size, cost, robustness etc? You might be able to see if the original was a 1/2 or 1 Watt resistor by its size. (2W and above are usually square instead of round, and white ceramic.) If it was a 1/2W resistor, then is that why it blew? Then again, it has lasted a long time. Alas I too know of big variations in car voltages. And as you see by the "squared" relationship, voltage changes have a BIG effect on power. (Hence for example a 20% increase in voltage from 12.0V to 14.4V means a 44% increase in power! [0.5W/0.43W is 1.16 => 1.07 increase in voltage (root of 1/16) => 1.08 x 14.4 = 15.5V. Remember I said I design for up to 16V (even though some vehicles can exceed that after cranking!)?] And we only have about 10% headroom in our Wattage.    What if the bulb is 6.6W? I think a 10% tolerance on light bulbs is not too conservative. Besides, with a 1W resistor, you could use 10W bulb (if they are more common etc). Why the resistor? I assume the resistor is there to limit inrush current and hence extend lamp life (since 10% external resistance has little effect on brightness (unless halogen)), but maybe no. Maybe it is there to preserve incomes. If only the resistor failed to short-circuit mode so your light continues operating.... The final say? From experience I know what a pain the so called "constant voltage" 12V system is. Try designing for 8V to 16V operation - that's a 4:1 power dissipation ratio! Though that power dissipation range is overcome by modern "constant power" loads, there is still a 2:1 input current ratio from 8V to 16V. Even fusing becomes tricky - a 1,536W input amplifier is 192A/128A/96A at 8V/12V/16V respectively (ie, same 2:1 range as the input voltage). Then add 30%-whatever derating when driving in hot weather, high altitudes, etc. Even 10 or 12V to 16V is tricky. And how many batteries stay above 12V under high loads? So, 0.5W resistor for a 0.43W load.....? But with likely wiring and switching voltage drops etc - quite likely fine. (Not in my 1965 vintage car - it produces a modern 14.4V and had minimal distribution resistance.) So there endeth the ramble. Thank you for you question. Next...?

Took me twice to read and reread to fully understand everything, You woulden't happen to be a teacher would you? I like the the resistor in series explanation regarding the water flow and a facuet limiting flow.  I use to be a mechanic for many years(and no i was not a "Parts Changer" i did check the resistence,amps,volts,etc in parts before i made the assumption that it maybe the problem) so i understand relays and electrical curcuits to an extent but of course never had to calculate resistence nor any other curcuits using ohms law, books told me the answer to that. 

Wish i knew a bit more bought the indepth calculations but never needed that in all my time working on 12V DC systems not even in upgrades due to the fact i only used factory parts and most were fairly simple to install, as my father use to say keep it factory old.  That of course unless there was a factory modification that was needed to be done. 

As for your 1959 vintage i hope your not talking bout an British car with the banana positive is ground system yikes and the lucas wiring on jaguars talk about death trap seen more then one on a XJS set fire. 

? Anywho i actually just have one last question not regarding the light system but a resistor in general.  Does the color change the resistors value in ohms as well such as a white resistor with 4 band color and a light blue resistor with the same band colors or is the color only to show that it is flame proof-or whatever they may decide to make it.

The band colors describe the resistance value regardless of the body color. The body color usually describes a special characteristic of the resistor. Usually white is flameproof and blue is flameproof and fusible, both in a matte finish. Some manufacturer veer off with these colors too, I've seen flameproof in green; where green are usually inductors. When in doubt it is safer to replace with a flameproof and fusible resistor.This information is provided only as a reference.

All circuits should be verified with a digital multi-meter prior to making any connections.

With thanks to 91stt for the body color & firepoof & fusing stuff. (I come across them in TVs etc but not vehicles.) Nothing technical follows, just more of my ramble.... And answering an OP's minor question....

fiat1980spyder wrote: Took me twice to read and reread to fully understand everything, You wouldn't happen to be a teacher would you? I like the the resistor in series explanation regarding the water flow ....

Thanks! I'm glad you re-read, and that you then did understand more. I don't give answers per se. Instead I try to educate so people can catch their own fish. But this bores he crud out of the semi- & experienced. But I find the technique doesn't work on forums. The bored miss the finer details. The confused become more so. The impatient want instant knowledge else answer(s). I teach & educate fine in personal situations (whether individuals or large groups) and to any capability, but not on written forums. In person, when someone disagrees, I can engage and learn, or keep handing out the rope. It is obvious who is correct, else that both are etc. But in forums, I now longer bother - too many resort to the flaming, or take the literal rather than the context, introduce other situations etc. I recently even had one remove all his previous posts [he reckons he didn't like my attitude! Yeah - my attitude that if you interrupt with something that is wrong, I will try to find why the disparity, else explain why you are wrong. But no - this guy insisted that water that bypasses the faucet robs the faucet of its flow! (Keeping in mind with electricity, if the voltage (water pressure) is the same, the flow remains unaltered into and through the same restriction. And there were other deliberate (or otherwise) attempts to justify her/his replies]. Ooops - I'm point scoring again. Or rather, I'm robbing points from another.... (LOL) Despite the above point-scoring, no, I amn't a teacher - not in the formal sense - ie not formally qualified to teach, have not formally taught children etc (company executives excluded). [After discussions with a colleague, I think we have decided not to pursue educating via written forums. The benefiting audience is insufficient - there are better ways....] But water is a great model for electricity. But like all models, it is merely that - a model. It sometimes of often fails. And can be ambiguous (I have seen the same "bucket" used to descrive a capacitor and an inductor, yet they are electrical opposites in normal considerartions!). And models cannot be taken too literally - like the above "one parallel string robs another of current" - that only occurs if the supply (voltage) is current limited (which in practice a supply water pipe might be, but in electrics, each voltage (source) is considered in itself to be zero-impedance - ie, have pressure, but no water/current flow restriction INTO it's own pipe/component. But that's where water is used for the basics. When they are understood, it's easier using Ohms Law etc (water doesn't work!). Somehow you understand it - or do we just accept it? (You understand energy don't you? Why not tell me what it is...? Describe it.... simply....) And make sure you know when you are using a model... The number of people that think that an atom is a nucleus with orbiting electrons (like the sun & planets) - no way - that's a very old model. But it still works for most of us...