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I know that a car battery can provide 1000amperes for starting while other components only need a few amperes to work. I thought that loads "ask" the battery for some electricity and the battery obliges.

I just recently bought USB ports that slot perfectly into one of the empty dashboard slots. They came with a glass fuse that was already blown.

I went and took out the blown fuse and closed the circuit to energize it (parallel connection to cigarette lighter | 15amp fuse.)

I thought I had a basic understanding of electricity in cars and that the USB port would never draw more power than what it needed. I just wanted to see those LEDs light up! and then it got fried instantly - it didn't even light up. I just smelled burned electronics.

I guess it didn't pull more than 15amperes because the cigarette lighter circuit was still working. But why did it fry itself after being energized? I know that the cigarette lighter circuit pretty much works as if it is directly connected to the battery(right??) - there's no electronics that regulates current on it right?

Am I wrong that in a car, the loads pull the amperes and that it is not the battery pushing it?

Did the USB port circuit pull too many amperes and fried itself, or did the battery push the amperes that the USB port circuit couldn't handle it?

I'm very confused because my 220V AC plug in water heater is just a hunk of metal and it doesn't pull 1000ampere from the wall.

So again, do loads pull the amperes, or do power supplies push it?

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    \$\begingroup\$ The simple answer is that the supply offers a certain voltage, and the load decides how many amps it will take at that voltage. You get power by multiplying amps times volts. So 12V at 100 amps is 1200 Watts. But 10 Amps at 220V is 2200 Watts. I suspect this question will be closed. So I decided to answer in a comment. \$\endgroup\$ – mkeith Nov 8 '20 at 8:22
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    \$\begingroup\$ They came with a glass fuse that was already blown. This. There was apparently a reason why the fuse was blown in the first case. There is nothing preventing a faulty (or badly designed, or with protection means disabled) electronic device to "ask" for more current than it can handle (thermally usually). \$\endgroup\$ – Martin Nov 8 '20 at 9:06
  • \$\begingroup\$ If the USB port is designed for 12 volt input, but burned out immediately when connected to 12 volts, then it was defective. A defective device doesn't tell you much about how current and voltage behave. \$\endgroup\$ – JRE Nov 8 '20 at 9:06
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    \$\begingroup\$ Since the fuse was already blown, it failed a test at the factory, but was shipped anyway. \$\endgroup\$ – JRE Nov 8 '20 at 9:07
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    \$\begingroup\$ that is not what the question is asking \$\endgroup\$ – jsotola Nov 8 '20 at 17:24
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The common terminus is that the load pulls current. An analogy will help you understand:

Take a tire charged with (air) pressure. If you now make a hole into that tire, how much air is going to flow out of the tire? Well, it depends on how big the hole is. A small hole provides a large resistance for air to leak out, while there will be a high air flow when the hole is big. You can think of voltage being an "electron pressure" and the air flow as electrical current.

In the above analogy, would you consider the tire (or a compressor) to push the air or does the hole pull air out? Sure, the compressor pushes air in some kind, but essentialy, the hole size in combination with existing pressure is responsible for the air to flow.

You see, the terminology of pushing/pulling current is of limited help unless you are used to the concept. In the end, there is just the trinity between pressure, resistance and air flow.

In electronics, this is exactly what the ohmic law expresses.

For your case that means, that the battery is not responsible for the occured damage. There must be something wrong with the USB port. In special, the 5V regulator inside it appears the be malfunctioning and probably provides its load with full 12V. In our analogy: the whole (load) is not meant to get that high pressure and it will blow up (like popping a baloon with a needle).

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