Unknown meter and short circuiting

Discussion in 'General Electronics Chat' started by kohlrak, Apr 7, 2010.

  1. kohlrak

    Thread Starter New Member

    Apr 7, 2010
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    0
    Firstly, i have a meter that i don't know what it is. Perhaps the second question will help determine what it is...

    At this point, i'm at chapter 5 section 3 of the book, and i have some sort of beginners electronics kit. I lost the manual to it some time ago, and i'm trying to learn electronics from scratch. Anyway...

    I connected the 1.5v battery to the unknown meter. It goes off the scale. So, i throw a 4.7KΩ resistor on there, so it lands right about 10 on this unknown scale on the meter. I short circuit the thing over a "key" switch, so it drops down to a little over 5 (or 6 depending on whether you go by the blue line or black line) on the scale when i close the circuit. This leads to the first problem. Obviously this isn't an ohmmeter, but either an ammeter or voltmeter. Since, voltage is consistent across circuits, this can't be a voltmeter, because i'm seeing change.

    However, when i put the short over a 4.7KΩ resistor, it suddenly doesn't change on the meter when i press the key down. I swap the plugs on the resistors just to be sure that it isn't a bad resistor, and it's the same thing. Further curious, i move the short onto the 2.2KΩ resistor, and it still doesn't move the needle on the uknown meter. I test a bunch of other resistors, even trying the 100Ω resistor, and no change.

    So, first question, "What meter do i have?" Second question, "why does the needle move with a negligible resistance wired short but not with a 100Ω resistor added to the short?"
     
  2. kohlrak

    Thread Starter New Member

    Apr 7, 2010
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    Hm... Reading ahead and doing some testing (with much more sleep) i realize it's an ammeter. However, the problem still remains. How comes adding a resistor to the short suddenly makes the voltage equalize, but otherwise not equalize? Did i miss something in the reading?
     
  3. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Your top picture has the battery shorted so the voltage is zero (or as low as the resistance of your wires).
    The lower picture has the battery loaded with a 4700 ohm resistor so the voltmeter that you made reads the voltage of the battery.
     
  4. kohlrak

    Thread Starter New Member

    Apr 7, 2010
    14
    0
    Thank you for your response.

    It's not exactly a voltmeter: it's still an ammeter. I see what i'm doing wrong (thanks to a buggy circuit simulator program for the computer which provided parts that i don't have access to). For future reference, a better explanation of the situation and a solution:

    "The short" will refer to the short in the top circuit. "The meter" will refer to the meter in the top circuit.

    Circuit #1: Ammeter lower than expected
    -The short: <1Ω
    -The meter: 4.7kΩ

    Circuit #2: Ammeter reads 10 units (units are currently unknown and aren't important)
    -The short: 100Ω
    -The meter: 4.7kΩ

    Circuit #2: Ammeter reads 10 units (units are currently unknown and aren't important)
    -The short: 4.7kΩ
    -The meter: 4.7kΩ

    Circuit #2: Ammeter reads 10 units (units are currently unknown and aren't important)
    -The short: 470kΩ
    -The meter: 4.7kΩ

    Solution: Less than 1Ω results in sucking more amps than the battery can output. As such, the battery performance degrades, and thus the ampage correlation to voltage no longer applies. If i wouldn't've been too cheap to get a DMM for dealing with even small circuits such as this, I would've probably realized that the voltage is still the same throughout, but only the ampage is "stolen" from the meter's branch of the circuit.

    Lesson learned: Get a DMM and learn not to assume a program isn't helpful just because it keeps crashing (ktechlab).
     
  5. kkazem

    Active Member

    Jul 23, 2009
    160
    26
    Hi,
    First, I've got to say that most of what you wrote is unintelligible; but your ckt diagram helped to make the circuit clear anyway. Your question is not understandable, but I'll do my best to answer what I think you mean, or at least give you some insight into it. I don't know if you have an ammeter or not, you probably do, but her is what's going on in your two circuits. The top circuit should read zero on the meter if you truly have a short circuit across the 1.5V battery or source. But I believe you said you did have a reading, which would mean that you don't have a short circuit as you've shown in your top schematic. The bottom schematic splits the current roughly in half between the parallel 4.7K OHM resistor and the ammeter with another 4.7KOHM in series with it. This is due to an ammeter's low coil resistance, compared to the 4.7K. You should have somewhere around 300uA (yes, microamps) thru the ammeter and about the same thru the other parallel 4.7K for a total current draw of about 600uA from the battery. Since the ammeter does have some resistance, the current in that branch will be a bit less than the current thru the shunt 4.7K resistor. If this was a hobbyist kit to learn electronics, they usually come with microammeters. Now, if you put the battery right across the ammeter without any series resistor, I'm surprised you didn't damage it as it would peg the meter very hard due to very low circuit resistance. I guess that's all I can tell you.
    Regards,
    Kamran Kazem
     
  6. kohlrak

    Thread Starter New Member

    Apr 7, 2010
    14
    0
    Thank you for your time, kkazem.

    Yeah, it's difficult to explain without a drawing. Anyway, in the end, i realize that i have an ammeter. Apparently, the wires i'm using are very poor conductors (they're silver colored, so they're not copper), but still lower than 1Ω.

    And yes, the first thing i did when i got this as a wee lad was use it as a battery tester. It's not exactly broken, and the needle still is close to 0, so i assume the damage i did only affects the magnetism of the magnets (which wear over time anyway, since this is pretty old).
     
  7. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    An old fashioned voltmeter (with a swinging needle) used a low current ammeter in series with a current-limiting resistor, exactly like you have.
     
  8. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
    69
    All 'mechanical' meters - ie. moving coil, moving needle - are in effect ammeters.

    The full scale current may be very low, like to 25 or 50 micro amps, but they are still current operated.

    As voltmeters, they are used with a series resistor chosen so the total (resistor plus meter) resistance gives the correct current through the meter at the chosen full-scale voltage.

    By switching resistors, you switch voltage ranges.

    If you look at any multimeter that has a mechanical meter, you will find in it's specifications an 'ohms per volt' figure; typically 20,000.

    1/20000 = 0.00005 or 50 millionths (Ohms law, 1V / 20000 Ohms) so that means a 50uA meter is used.
    That allowed the user to calculate the load the meter would put on the circuit being tested for any selected voltage range.
     
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