I have a somewhat vague understanding on wattage and voltages but for lamps, if you were given a lamp with 45w and put a 60w electric bulb in it, doesn’t it follow that it will just power 45w of the 60w and produce a weaker power, less heat, and a dimmer light; and if placed the other way around(60w lamp/45w bulb), the lamp will provide more than the bulb can handle and the possibility of a broken bulb is imminent?
But a little online search says that it’s the opposite. It’s confusing especially all I see online are from bulbs connected to an alternating current. Does it only apply to bulbs powered by AC or does it apply to DC as well like bulbs found in cars?
In: Engineering
The lamp, which will have all the wiring set up to withstand the heat and amperage of a 45 watt bulb, will overheat if a 60 watt bulb is installed over time. A lamp is, simply put, a pretty case to hold wires and the bulb, it in no way limits the flow of electricity itself. The wires do. The Load in this case the bulb will demand the wattage from the source, and the system has to withstand the forces.
If you put a 45 watt bulb in one that is set up for 60 watts the system is over engineered so to speak.
There are other factor, but this is the most simple explanation I can think of at the moment.
Watts = volts * amps
So long as the voltage is right the bulb should be fine.
If the lamp is rated 45w and the bulb is rated 60, that means the lamp will possibly melt or have a fault if you put a bulb of more than 45w in it. Best case it’ll blow a fuse, worst case a fire.
Bulbs made for AC generally won’t work on DC. Check the type of bulb first. Generally though the voltage is the most important – a 9v supply won’t light a bulb intended for 230v, whereas a 230v supply for a 9v bulb will give a small explosion.
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