Why does circuit break when jumper inserted to power?

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When a jumper wire is connected directly from the positive to the negative terminal of a battery, it creates a short circuit, causing the current to bypass the LED, which results in the bulb turning off. This occurs because the internal resistance of the battery limits the voltage available to the LED, leading to insufficient power for it to light. In contrast, using a voltmeter or a switch does not create the same effect because they do not draw as much current, allowing the LED to function properly. The discussion emphasizes the importance of understanding Ohm's Law and the concepts of voltage, current, and resistance in circuit design. Ultimately, short-circuiting a battery can be dangerous and should be avoided to prevent damage or hazards.
  • #31
Are you sure the other measurements will still work if the fuse is blown? I opened it up and it doesn't look like anything is blown.

To reiterate as I mentioned above, the meter still works. I get measurements for Ohms and V. It's only Current that sits at 0 no matter what I do.
 
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  • #32
dbconfession said:
Are you sure the other measurements will still work if the fuse is blown? I opened it up and it doesn't look like anything is blown.

To reiterate as I mentioned above, the meter still works. I get measurements for Ohms and V. It's only Current that sits at 0 no matter what I do.

most likely other measurements will still work and you answered your own question ;)

ohhh BTW NEVER put a meter across a battery or other power source when in Ohms range
It isn't good for the meter

depending on the style of fuses in there, they may blow without any obvious look as such

IF the Ohms range is still working, remove the fuses in the Amps/mA ranges that VK6KRO spoke of and use the Ohms range to test them ... you should get 0 (zero) Ohms ... if the meter reads OL then the fuse(s) have blownDave
 
  • #33
Thanks Dave. I can't keep my eyes open. Lol. I'll try in the morning. Can you draw me a simple breadboard circuit to test current with a 9V? Reading the manual I realize that testing current requires the circuit to be broken first and then closed with the multimeter. I want to make sure I have the proper setup and know how much the current should read with that setup so I can confirm that I'm getting the same reading. Thanks again.
 
  • #34
You can calculate a current using Ohm's Law if you know the voltage across a known resistance.

Current = Voltage divided by resistance in ohms.
A 9 volt battery will comfortably give 20 mA. If you could get a 470 ohm resistor and wire it across a 9 volt battery, this would give a current of about 20 mA.

Current = 9 volts / 470 ohms = 19.14 mA.

This is a bit uncertain, because the battery voltage will be more than 9 volts if it is fairly new.

It would be good enough to tell if the meter was measuring current or not.
 
  • #35
db
there's a simple memory aid that everybody learns the hard way. It becomes so natural that we forget to properly emphasize it to beginners.

An ammeter's job is to measure current. So naturally it doesn't impede the flow of current, instead it wants current to flow through itself with the greatest of ease. That way it doesn't affect the circuit you're testing.

That's logical enough. But here's the corollary:
If you hook an ammeter straight to a voltage source, what limits the current ? Not the ammeter, it welcomes current.
So the voltage source will push through that unlucky ammeter every last drop of current that it can produce.
A fresh 9v battery can deliver a half amp for a very shot while and wreck a sensitive meter. Or blow a 1/4 amp fuse.

So the thing to remember is this:
" Ammeters are pushovers."
That short phrase should remind you of this rule:
NEVER connect an ammeter where there's not something else in the circuit that limits the current.
Ammeter ALWAYS goes in series with the load you are testing.
Set it for the highest amp range and switch to a lower one after you're sure current isn't way more than you expected it to be.

Now you know why.
Stock up on meter fuses. I do.

Practice question: How will you connect your meter and vk6r0's 470 ohm resistor to make his suggested check?

old jim
 
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  • #36
Kavik said:
Here is a table for wire gauge (AWG) and Ohms (resistance) per 1000 feet. The only columns you need to worry about are the first and third. You can see that 1000 feet of a 22 gauge wire has a resistance of 16.0 Ohms. I would say a good estimate for the length of your wire is 1 foot. This gives a total resistance for your 22 gauge wire of 0.016 Ohms:

{\frac{16.0 \, Ohms}{1000 \, feet}} = 0.016 \, Ohms.

http://www.picwire.com/images/paper9table1.gif
Source: http://www.picwire.com/technical/wire.php

Kavik. You'd posted this chart for copper wire reference. Some of the wire I'm using that came with a radio shack learning kit is visibly not copper (color of bare wire is silver). Do you think it's stats are comparable to those in your chart? And do you know what material is likely being used?
 
  • #37
It is probably tinned copper. Tinned copper is a bit easier to work with when soldering so I could easily see it being included in any learning kit. It is also less prone to corrosion.

Any differences in stats would be negligible for what you're doing since it's mostly copper. Try looking at the end of the wire after cutting it, you might be able to see a dot of copper in the center of the end's cross section (easier to see on solid core or larger wire). Scratching the wire down to the core might reveal the copper color as well.
 

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