What does 12v 36w mean on a bulb?

  • Thread starter Thread starter fatoomch
  • Start date Start date
  • Tags Tags
    12v Bulb Mean
AI Thread Summary
The discussion clarifies that a bulb rated at 12V and 36W will consume 36 watts of power when connected to a 12V supply, drawing 3A of current. The '12V' designation indicates optimal performance and longevity at that voltage. Connecting the bulb to a 24V supply will result in it consuming 72 watts, leading to rapid burnout due to excessive heat. The relationship between voltage and power consumption is explained by the formula P = V^2/R, indicating that doubling the voltage quadruples the power. The resistance of the filament increases with temperature, complicating the expected outcomes when altering voltage.
fatoomch
Messages
8
Reaction score
0
Dumb question, I know.

But what exactly does it mean? Does it mean 36watts is what a bulb will consume at 12v?.. And does '12v' mean that the bulb will only work at 12v. What if you connect the bulb to a 24v supply?

Please excuse my ignorance.
 
Engineering news on Phys.org
At 24V it will glow much brighter, and for a much shorter time.


BTW, a 36w bulb with 12v will pass 3A. That's huge. This must be in a car.
 
Yes, your bulb will draw 36 Watts (or Joules per second) when connected to a 12V power supply.
 
ok thanks... so when it states '12v' it just means this bulb will work best (last longer) and is specifically designed for a 12v supply?

...And if it IS connected to 24v supply it will consume 72watts of power?
 
It's designed for 12V. It will probably burn itself out very rapidly if you double the voltage, as you'd be subjecting its filament to four times as much heat dissipation.

Power goes up with the square of the voltage (or the square of the current):

P = I^2 R = \frac{V^2}{R}

If you double the voltage, you quadruple the power consumed.

- Warren
 
However, reality is not always this simple. One of the phenomena explored in a later chapter is that of conductor resistance changing with temperature. In an incandescent lamp (the kind employing the principle of electric current heating a thin filament of wire to the point that it glows white-hot), the resistance of the filament wire will increase dramatically as it warms from room temperature to operating temperature. If we were to increase the supply voltage in a real lamp circuit, the resulting increase in current would cause the filament to increase temperature, which would in turn increase its resistance, thus preventing further increases in current without further increases in battery voltage.

http://www.allaboutcircuits.com/vol_1/chpt_2/5.html"
 
Last edited by a moderator:
Hi all I have some confusion about piezoelectrical sensors combination. If i have three acoustic piezoelectrical sensors (with same receive sensitivity in dB ref V/1uPa) placed at specific distance, these sensors receive acoustic signal from a sound source placed at far field distance (Plane Wave) and from broadside. I receive output of these sensors through individual preamplifiers, add them through hardware like summer circuit adder or in software after digitization and in this way got an...
I have recently moved into a new (rather ancient) house and had a few trips of my Residual Current breaker. I dug out my old Socket tester which tell me the three pins are correct. But then the Red warning light tells me my socket(s) fail the loop test. I never had this before but my last house had an overhead supply with no Earth from the company. The tester said "get this checked" and the man said the (high but not ridiculous) earth resistance was acceptable. I stuck a new copper earth...
I am not an electrical engineering student, but a lowly apprentice electrician. I learn both on the job and also take classes for my apprenticeship. I recently wired my first transformer and I understand that the neutral and ground are bonded together in the transformer or in the service. What I don't understand is, if the neutral is a current carrying conductor, which is then bonded to the ground conductor, why does current only flow back to its source and not on the ground path...
Back
Top