"O" level physics question - electrickery!
Discussion
I've just about retained/remembered Ohms Law and the related relationsips/formulae between Volts, Amps, Watts & Ohms, but...
I was trying to work out what happens (in terms of perceived "brightness") if you replace your 12V 60/55W headlamp bulbs directly with those designated as 120/110W - but for a 24V system?
Do they still operate as 120/110s...or double that...or half that...or what? And, leaving aside the mathematical relationship and the wattage, do they APPEAR the same, brighter or duller than the ones they replace? I thought that I might have found a cheap lighting upgrade; but maybe it's not that simple. (Life rarely is!)
My head hurts!
I was trying to work out what happens (in terms of perceived "brightness") if you replace your 12V 60/55W headlamp bulbs directly with those designated as 120/110W - but for a 24V system?
Do they still operate as 120/110s...or double that...or half that...or what? And, leaving aside the mathematical relationship and the wattage, do they APPEAR the same, brighter or duller than the ones they replace? I thought that I might have found a cheap lighting upgrade; but maybe it's not that simple. (Life rarely is!)
My head hurts!
To early in the morning for me to bother working out the maths of it BUT
The 24v (lamps) NOT bulbs piff will be duller because the resistance of the
filament is roughly twice what it would be for a 12 V system so
Half V for twice R = a lot less I
So lower temp in the wire and fewer photons emitted maybe
ET reinsert a line of text DOH!
The 24v (lamps) NOT bulbs piff will be duller because the resistance of the
filament is roughly twice what it would be for a 12 V system so
Half V for twice R = a lot less I
So lower temp in the wire and fewer photons emitted maybe
ET reinsert a line of text DOH!
Edited by FranKinFezza on Sunday 17th April 01:49
AdvanceRoadcraft said:
I've just about retained/remembered Ohms Law and the related relationsips/formulae between Volts, Amps, Watts & Ohms, but...
I was trying to work out what happens (in terms of perceived "brightness") if you replace your 12V 60/55W headlamp bulbs directly with those designated as 120/110W - but for a 24V system?
Do they still operate as 120/110s...or double that...or half that...or what? And, leaving aside the mathematical relationship and the wattage, do they APPEAR the same, brighter or duller than the ones they replace? I thought that I might have found a cheap lighting upgrade; but maybe it's not that simple. (Life rarely is!)
My head hurts!
Well, we need to know what the power of a 24V lamp is in a 12V system. This is ignoring a VERY important property of halogen lamps, specifically that when cold the filament is very low resistance. It is a "non ohmic" conductor. So as it gets hotter the resistance actually goes up. The emission is non-linear too, as a cooler filament may tend to emit more in the heat / infra-red band rather than the visible spectrum.I was trying to work out what happens (in terms of perceived "brightness") if you replace your 12V 60/55W headlamp bulbs directly with those designated as 120/110W - but for a 24V system?
Do they still operate as 120/110s...or double that...or half that...or what? And, leaving aside the mathematical relationship and the wattage, do they APPEAR the same, brighter or duller than the ones they replace? I thought that I might have found a cheap lighting upgrade; but maybe it's not that simple. (Life rarely is!)
My head hurts!
Anyway, ignoring that...
Current = Power / Volts
= 110W / 24V
= 4.58A (at 24V)
Resistance = Voltage / Current
= 5.24 Ohms
Combining Ohms and Power laws,
Power = V^2 / R
= 27.5 Watts (Half the power of a 50/55W)
In other words...
In a linear system, power squares as the voltage. So double the voltage and the power goes up by a factor of 4. Conversely, halve it and the power drops to a quarter.
Human eye brightness perception is logarithmic, so the perceived brightness is approximately 60%. This is just the input power though, a greater proportion of the power emitted may well be in the infra-red band, as the filament is running cooler.
Zad said:
Well, we need to know what the power of a 24V lamp is in a 12V system. This is ignoring a VERY important property of halogen lamps, specifically that when cold the filament is very low resistance. It is a "non ohmic" conductor. So as it gets hotter the resistance actually goes up. The emission is non-linear too, as a cooler filament may tend to emit more in the heat / infra-red band rather than the visible spectrum.
Anyway, ignoring that...
Current = Power / Volts
= 110W / 24V
= 4.58A (at 24V)
Resistance = Voltage / Current
= 5.24 Ohms
Combining Ohms and Power laws,
Power = V^2 / R
= 27.5 Watts (Half the power of a 50/55W)
In other words...
In a linear system, power squares as the voltage. So double the voltage and the power goes up by a factor of 4. Conversely, halve it and the power drops to a quarter.
Human eye brightness perception is logarithmic, so the perceived brightness is approximately 60%. This is just the input power though, a greater proportion of the power emitted may well be in the infra-red band, as the filament is running cooler.
Well put Zad i was trying to keep it simpler than that by just showing (in a very rough way) that if the lamp is expecting to see twice the voltage then its only goingAnyway, ignoring that...
Current = Power / Volts
= 110W / 24V
= 4.58A (at 24V)
Resistance = Voltage / Current
= 5.24 Ohms
Combining Ohms and Power laws,
Power = V^2 / R
= 27.5 Watts (Half the power of a 50/55W)
In other words...
In a linear system, power squares as the voltage. So double the voltage and the power goes up by a factor of 4. Conversely, halve it and the power drops to a quarter.
Human eye brightness perception is logarithmic, so the perceived brightness is approximately 60%. This is just the input power though, a greater proportion of the power emitted may well be in the infra-red band, as the filament is running cooler.
to illuminate to about half its rated brightness.
Any how do halogen lamps (automotive quality) have a linear response?
...I've got it now.
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