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There are hundreds of different types of light bulbs that we can use. Some of these types of bulbs are for everyday usage and will last for months before we have to change them out. While others are a bit more expensive and only used for decorative purposes. One of the most popular types of these are the Christmas mini lights.

The mini lights were first created during the 1970’s and since that time have become the most popular types of strand lights. They have a voltage of 2.5 and use incandescent light bulbs. These look similar to the kind that are used inside of flashlights. Many of us will use them during the holiday season – like Christmas – or for parties or weddings.

One of the biggest things that made these so popular was their ability to light up a room with a special glow and the way that they were able to work. Many have wondered how something that is only 2.5 volts is able to work in an outlet that is 120 volts. The only way that they can work properly is if they are connected in a group or series. If you have 48 bulbs with each emitting 2.5 volts then you have 120 volts plugged in at once.

Still many mini strand lights are built with a total of 50 bulbs. This enables them to work properly and to give them a better chance to work even if one goes out. However, sometimes the extra two do not work that way. This explains why when one of the lights on the Christmas tree go out they all seem to stop working.

At least that is how it was when they were first created. The going out of one single light bulbs would affect all and the whole thing would cease to work. But now it can burn out and the other will still be able to work well without it. The only time everything will stop working is if it is taken out of its socket. It has the ability to do this through a shunt wire.

The shunt wire is wrapped around two separate posts that are within the mini light bulb. This wire has been built with a coating that makes it resistant until the filament is going to fail. When this happens there will be heat emitted from it – which will cause a current to travel through the shunt. This in turn will burn the coating off and thus reduce the resistance that it once had.

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