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A question regarding fuses.

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gonadgranny

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Hi. I have constructed a circuit that consists of three 40 watt bulbs. They will be running off a 240v mains power supply.
When running a single bulb they require a 3amp fuse. I am thinking that now there are three bulbs I need to change the fuse.
Will putting a higher load fuse in cause any other problems? I don't need to wire up a separate fuse for each bulb do I?
Oh, they are wired in parallel.
Thank you very much in advance to anyone who responds.
Danny.
 

a 40W bulb uses P=V*I => I=P/V =40/220=0.182A ,
3*0.182A= 0.546A
with you current fuse (3A) you can use more than a dozen of bulbs without any problem.

Alex
 
3 x 40Watt bulb = 120Watts

Amps = Watts / Volts

120 / 240 = 0.5 Amps

No, I don't expect your 3 Amp fuse is going to have any difficulty in passing the 0.5 Amps for the 3 light bulbs.
 
What the purpose of the fuse in your circuit? In can be either overcurrent or short circuit protection. The former would be required, if there's a risk that cables, switches, connectors would overheat when the load consumes more current than expected. Normally, this must not be expected for an incandescent lamp. Short circuit protection is normally provided by the switchboard for all electrical circuits in your house and won't be necessarily required in most cases.
 
nice on chaps!
what helpful little forum this is. The only fuse is the one in the original plug and after the very helpful replies it seems i won't need any more!
Cheers fellas.
Danny.
 

Danny,

Unfortunatly you will need more help.

The first question is what type of "40Watt" bulbs are they?

It's just that fillement bulbs and quite a few others have a rating for when the are hot not when they are cold. The filement of a bulb has very low resistance when cold and sometiemes as much as ten times the resistance when at the working temprature. To get there the bulb needs considerably more power and you get an "inrush current". It is this that used to cause the very anoying problem that old style filement bulbs had a habit of "blowing" when turned on not during normal operation.

Thus you have to consider not just the "hot current" capacity of your bulbs but the "cold inrush current" of your bulbs and use a "slow blow" fuse etc.
 

Incandescent light bulb - Wikipedia, the free encyclopedia

"Electrical characteristics
Incandescent lamps are nearly pure resistive loads with a power factor of 1. This means the actual power consumed (in watts) and the apparent power (in volt-amperes) are equal. The actual resistance of the filament is temperature-dependent. The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating. For example, a 100-watt, 120-volt lamp has a resistance of 144 ohms when lit, but the cold resistance is much lower (about 9.5 ohms).[33][64] Since incandescent lamps are resistive loads, simple triac dimmers can be used to control brightness. Electrical contacts may carry a "T" rating symbol indicating that they are designed to control circuits with the high inrush current characteristic of tungsten lamps. For a 100-watt, 120-volt general-service lamp, the current stabilizes in about 0.10 seconds, and the lamp reaches 90% of its full brightness after about 0.13 seconds.[65]"

Fuse (electrical) - Wikipedia, the free encyclopedia
"A standard fuse may require twice its rated current to open in one second, a fast-blow fuse may require twice its rated current to blow in 0.1 seconds, and a slow-blow fuse may require twice its rated current for tens of seconds to blow."

Do you think that 100mS will be enough to blow his fuse?

Alex
 

I agree to your assumption, that a standard F3 fuse won't be blown by a 3x40W incandescent lamp load. However, If you are no designing a professional product, you may risk the trial instead of doing lengthy calculations.
 

Another thing to remember: Fuses are rated at the current they can carry without blowing, not the current at which they will blow. Also, make sure that the fuse current rating is not higher than the current carrying capacity of the conductors being used.
 
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