I want to dim light bulb by 50% I want to learn how to calculate it on my own

[Omron]

I have a 5W lamp on 12Volts battery, I want it to be half as bright, so I will add a resistor. What resistor should I supply?

A = W = 5 = 0.416mA
_ _
V 12


R = V = 12 = 28.84 Ohms (R) this is resistance of lamp
_ _____
I 0.416

So I will add one resistor 28.84 Ohms

I will have 28.84 + 28.84 = 57.6 Ohms

Or 28.84 + 28.84 = 14.42 Ohms

I don't know if it will work in series or in parallel
voltage would be 6V and amp current mA = 0.208mA We divide everything by two

W= V * I = 6 * 0.208 = 1.248Watts resistor would be needed and 28.84 Ohms

Is my calculation correct? And I need explanation why its divided by two I'm little confused

[momaka]

Is my calculation correct? And I need explanation why its divided by two I'm little confused

Your calculations are correct...
The approximate resistance of light bulb will indeed be: P = V^2 / R... or R = V^2 / P = 12^2 / 5 = 144 / 5 = 28.8 Ohms.

But here is where you got tricked: you probably assumed that if you use a 28.8 Ohm resistor in series with the bulb, the voltage going to the bulb will be divided in half (6V), which is CORRECT. However, the power usage of the bulb will not be divided in half...

For pure resistive components, dropping the voltage in HALF (1/2) means their power usage will go down to a QUARTER - i.e. 1/4 of what it was.
What's interesting is that you even found this with your own calculations, which showed that the bulb will be using 1.248 Watts, which is CORRECT again, as 1.25 Watts is 1/4 of 5 Watts.

Now, to make the bulb use only HALF of its rated power (i.e. 2.5 Watts), you can use this formula P = I^2 * R, where we use 2.5 Watts for P and 28.8 Ohms for R to get the current, I, going through the bulb. So...
I = sqrt(P / R )... or = (P / R )^0.5 = (2.5 / 28.8)^0.5 = (0.086805555)^0.5 = 0.29463 Amps

Next, we calculate the voltage across the bulb with this current flowing through it. To do that, we simply use V = I * R = 0.29463 * 28.8 = 8.49 Volts
Now, we still don't know the value of the resistance of the resistor to use, but we do know it has to pass approximately 0.295 Amps through it and there must be 12 - 8.49 = 3.51 Volts across it.
So to get the value of the resistor we need, the following equation can be used again: R = V / I = 3.51 / 0.295 = 11.9 Ohms ... If you don't care to be driving the light bulb at *exactly* half of its rated power, then you can go up or down a bit on that resistance, so you can have an easier time finding a standard value resistor. In your case, both a 10 Ohm or a 12 Ohm resistor will give reasonably close results.

But we are not done yet! There's one more calculation: to see how much power (heat) the resistor will dissipate. If you pick a resistor with too low of a power rating, it will burn out quickly.
Here's how we calculate the power rating of the resistor: we already know the current going through it and we know its resistance (about 12 Ohms, so let's go with a 12 Ohm standard value).
P = I^2 * R = 0.295^2 * 12 = 1.04 Watts .... so a 2 or a 3 Watt -rated resistor will be good to go (it will still run a little hot, though.)

On the other hand, if you use a 10 Ohm resistor, you can't really use the above equation anymore, since the current won't be 0.295 Amps anymore. It will be a bit higher. But that's easy to calculate now:
I = V / R = 12 / ( 10 + 28.8) = 12 / 38.8 = 0.3093 Amps
The bulb will then use about P = I^2 * R = 0.3093^2 * 28.8 = 2.76 Watts
And the 10 Ohm resistor will be dissipating about P = I^2 * R = 0.3093^2 * 10 = 0.95666 Watts... so a 2 or 3-Watt rated resistor will be needed again so it doesn't run as hot.

Hope that was enough math for you for the day.

Alternatively, just get one of those cheap $ step-down buck regulators found on eBay or AliExpress and turn down the voltage going to the bulb to whatever is needed to get the desired bulb brightness (or rather, dimming.)
As harp noted above, half the output power doesn't necessarily translate to half of the output lumens - at least not for incandescent / halogen bulbs. So you will need to experiment a little too, to get the brightness / dimming you want.

[Omron]

https://ohmslawcalculator.com/voltag...der-calculator
https://ohmslawcalculator.com/ohms-law-calculator
By this calculator it seems like I calculated well and that resistor needs to be put in is indeed 28~ Ohms...Two resistors in parallel in fact so I have 14.4 Ohms~
And about watts it would be
Ampers = Watts = 5 = 0.83miliAmps
_____ _
Voltage 6

Watts = Voltage * I (Ampers) = 6 * 0.83 = 4.98 watts

So my resistors needs to be two resistors in parallel 28.8 Ohms each, which would give me 14.4 Ohms resistance and I need 5 watts resistors to be able to take this current... Need confirmation

[petehall347]

the lamp resistance will change when its powered up .

[harp]

I want it to be half as bright, so I will add a resistor.

You are on wrong track, brightness is not linear with resistance.
Say, if you have 100w lamp, and in series add one more 100w lamp, result of each one brightness is less than one single 50w lamp.

[Omron]

So what should I do?

[stj]

pwm control
you can make a circuit with an 8pin chip and a mosfet or just buy a module designed for 12v motor/fan control from aliexpress for a $
if you want to make one, the chip i would use is a 555 - read the datasheet to see how usefull it is!

if you use a 555 and a fet you wont be burning lots of energy as heat, the circuit will just be switching the lamp very fast to control the brightness.

[petehall347]

you might be best to see how a 3w lamp looks

[harp]

The brightness of light (lumens) you can measure with some mobile aplication like light meter, so you have reference point on specific location, say table.
When that divide in half, that is what you need.

If the heat of resistors is not a problem, you can build a simple resistor network to experimentaly adjust needed value, and able to set it precisely, example in attachment. Do not use potenciometer if not rated for this aplication, or you are not sure about outcome.
The circuit with transistor (on heatsink) allow you to use ordinary low power potenciometer to control brightness, I guess that you want analog solution...

[Omron]

Well I actually wanted to make this bulb dimmer, its shining into eyes during night, its on tractor vehicle, I thought that adding some sort of resistor would do the job, since bulb is sealed inside, I didn't wanted to take it apart and to put smaller one inside, but once it dies I will do it

[stj]

if its for a vehicle, just get a ReoStat
go to a scrapyard and you can find them on most old (older than 80's) cars

[petehall347]

stick a sheet of lighting gel over it

[sam_sam_sam]

if its for a vehicle, just get a ReoStat
go to a scrapyard and you can find them on most old (older than 80's) cars

Or get a new car with a digital PWM controller that is made for 12 volt LEDs or incandescent light bulbs the majority of them work very well however you need to trace the wires of the circuit so you know how to hook it up correctly

[Omron]

Well thx guys I think I have enough ideas now

[momaka]

Is my calculation correct? And I need explanation why its divided by two I'm little confused

Your calculations are correct...
The approximate resistance of light bulb will indeed be: P = V^2 / R... or R = V^2 / P = 12^2 / 5 = 144 / 5 = 28.8 Ohms.

But here is where you got tricked: you probably assumed that if you use a 28.8 Ohm resistor in series with the bulb, the voltage going to the bulb will be divided in half (6V), which is CORRECT. However, the power usage of the bulb will not be divided in half...

For pure resistive components, dropping the voltage in HALF (1/2) means their power usage will go down to a QUARTER - i.e. 1/4 of what it was.
What's interesting is that you even found this with your own calculations, which showed that the bulb will be using 1.248 Watts, which is CORRECT again, as 1.25 Watts is 1/4 of 5 Watts.

Now, to make the bulb use only HALF of its rated power (i.e. 2.5 Watts), you can use this formula P = I^2 * R, where we use 2.5 Watts for P and 28.8 Ohms for R to get the current, I, going through the bulb. So...
I = sqrt(P / R )... or = (P / R )^0.5 = (2.5 / 28.8)^0.5 = (0.086805555)^0.5 = 0.29463 Amps

Next, we calculate the voltage across the bulb with this current flowing through it. To do that, we simply use V = I * R = 0.29463 * 28.8 = 8.49 Volts
Now, we still don't know the value of the resistance of the resistor to use, but we do know it has to pass approximately 0.295 Amps through it and there must be 12 - 8.49 = 3.51 Volts across it.
So to get the value of the resistor we need, the following equation can be used again: R = V / I = 3.51 / 0.295 = 11.9 Ohms ... If you don't care to be driving the light bulb at *exactly* half of its rated power, then you can go up or down a bit on that resistance, so you can have an easier time finding a standard value resistor. In your case, both a 10 Ohm or a 12 Ohm resistor will give reasonably close results.

But we are not done yet! There's one more calculation: to see how much power (heat) the resistor will dissipate. If you pick a resistor with too low of a power rating, it will burn out quickly.
Here's how we calculate the power rating of the resistor: we already know the current going through it and we know its resistance (about 12 Ohms, so let's go with a 12 Ohm standard value).
P = I^2 * R = 0.295^2 * 12 = 1.04 Watts .... so a 2 or a 3 Watt -rated resistor will be good to go (it will still run a little hot, though.)

On the other hand, if you use a 10 Ohm resistor, you can't really use the above equation anymore, since the current won't be 0.295 Amps anymore. It will be a bit higher. But that's easy to calculate now:
I = V / R = 12 / ( 10 + 28.8) = 12 / 38.8 = 0.3093 Amps
The bulb will then use about P = I^2 * R = 0.3093^2 * 28.8 = 2.76 Watts
And the 10 Ohm resistor will be dissipating about P = I^2 * R = 0.3093^2 * 10 = 0.95666 Watts... so a 2 or 3-Watt rated resistor will be needed again so it doesn't run as hot.

Hope that was enough math for you for the day.

Alternatively, just get one of those cheap $ step-down buck regulators found on eBay or AliExpress and turn down the voltage going to the bulb to whatever is needed to get the desired bulb brightness (or rather, dimming.)
As harp noted above, half the output power doesn't necessarily translate to half of the output lumens - at least not for incandescent / halogen bulbs. So you will need to experiment a little too, to get the brightness / dimming you want.

[sam_sam_sam]

—>Hope that was enough math for you for the day. <—

I always like reading your explanations of how to figure out ohms law calculations and how to troubleshoot devices keep up the good work
I have learned a lot from your posts

[Omron]

Thank you momaka for explaining formulas to me, and to everyone else who was willing to help me.
I will sit down with piece of paper and calculator and do all calculations by myself again, its easier to learn once you dip the paws in how they say hehe

[Omron]

I put it on paper, easier to calculate that way, by the way my notification lamp is connected at same fuse as my headlights and headlight switch.

Of course it turns on lights when I turn on lights on machine, don't want it to be running notification light non stop without being able to switch them off of course..

My question was what if I put like 5watt 12 Ohm rsistor? Will it do just as fine job as 3watt 12 Ohm resistor? maybe slightly better with heat dissipated?

Also, since more lamps are connected in this circuit, will they affect my Current drawn etc, like we did calculus for one lamp, but do we need to take into consideration all lamps? like few more oil lamps and headlight notification lamp and also headlights rear lights, blinkers themselves, all in one circuit..

[sam_sam_sam]

I put it on paper, easier to calculate that way, by the way my notification lamp is connected at same fuse as my headlights and headlight switch.

Of course it turns on lights when I turn on lights on machine, don't want it to be running notification light non stop without being able to switch them off of course..

My question was what if I put like 5watt 12 Ohm rsistor? Will it do just as fine job as 3watt 12 Ohm resistor? maybe slightly better with heat dissipated?

Also, since more lamps are connected in this circuit, will they affect my Current drawn etc, like we did calculus for one lamp, but do we need to take into consideration all lamps? like few more oil lamps and headlight notification lamp and also headlights rear lights, blinkers themselves, all in one circuit..

Yes if you want them to be dim as well otherwise no

[Omron]

I only want my notification bulb to be dimmed, so what do I do, maybe best option would be to switch lower watt bulb, it will draw less current and less lumens produced

[sam_sam_sam]

I only want my notification bulb to be dimmed, so what do I do, maybe best option would be to switch lower watt bulb, it will draw less current and less lumens produced

Or higher working voltage light bulbs will also cause them to appear dimmer than a normal voltage bulb ran at the right voltage

Here is an example let say a you have a 24 volt power source and you use a bulb rated for 28 volts is going to appear dimmer than a light bulb that is rated for 24 volts

So let say that you were able to find an incandescent bulb that is rated for 18 volts or even 24 volts might work or it might be too dim