Vehicle electric lesson needed!!

Windmill John

Windmill John

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I have a degree in over complicating things and I'm sure this will be no different!
I was just fiddling and came across my question.
As you can see, series circuit, two bulbs. Let's cut to the chase. Connect to battery negative and only the 2 watt bulb lights; they do work independently.
Now, this could be a perfect example of voltage drop where we pretend the 2 watt bulb is corrosion etc. stopping the 'load' (21 watt bulb) from lighting.
Question time and I'm sure it relates to Ohm's law.
Why did the 21 watt bulb not light at all?
21 watt bulb by itself would be about 0.5 amp using Ohm's.
2 watt bulb by itself would be about 6 amp using Ohm's
Of course I could be wrong.
Bulbs in series (possibly confusing myself now) 21 watt + 2 watt, 23 watt, V=IR. about 0.5 amp.

If that is all correct, why didn't the 21 watt bulb light even a bit?
Is it due to the 2 watt bulb pulling far more current and therefore not enough current to light the 21 watt? If that is correct, I understand it all and have no question.
If not correct, please explain as it might drive me crazy. :rolleyes:
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circuit.jpg
 
Well, thats quite simple.
You have two bulbs with resistances (21W | 6,85 Ohms; 2 W | 72 Ohms).
Bulb 21W draws 1,75 amps from battery (connected directly), Bulb 2W - 0,17 amps.
As they are connected in series total current in the system is 12V/(72+6,85 Ohms)), so common current for bulbs is 0,15 amps. Thats 45 times less than required.
 
Note that much of Europe uses a comma where a decimal point is used in the English system, i.e., 6,85 is the same as 6.85, etc.

And, yes, John, it's just Ohm's law as Andrew explains nicely. The low-Watt bulb restricts current through both bulbs (in series) to the point that the filament on the high-Watt bulb does not receive enough juice.

I think our motorcycles keep bulbs in parallel.
 
Whoops, just seen my glaring mistake!! I used Ohm’s law equation using power instead of resistance. Both your posts make total sense and appreciate that.
Thank you for the replies. Piece of cake this electrics debacle 😊
And yes AndrewJK, it is simple when I switch on my head.
Thanks stl360+450, aware of the series, parallel issue, think back to early Christmas lights 😉
 
Whoops, just seen my glaring mistake!! I used Ohm’s law equation using power instead of resistance. Both your posts make total sense and appreciate that.
Thank you for the replies. Piece of cake this electrics debacle 😊
And yes AndrewJK, it is simple when I switch on my head.
Thanks stl360+, aware of the series, parallel issue, think back to early Christmas lights 😉
 
If you're not already aware, the power formula for direct current is P=I*V. Ohm's law says V=I*R or I=V/R, so you get some new expressions for power by substitution:
P = I * V = I^2 * R = V^2 / R.

So, you can get the resistance of a bulb by computing R = V^2 / P.
 
Thanks. That does throw me a bit I’m afraid. I just did it the John way!
P=VI, work that out to get the current, then do V=IR.
Sorry I’m quite simple in these things, probably explain why carbs etc are easier…
I’ll sit down with a beer later and look at your equation.
With some equations, I think a wall just goes up in my head.
 
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