Noob Question: Measuring the current in a voltage divider circuit

Discussion in 'General Electronics Chat' started by donkeyrobber, Aug 13, 2014.

  1. donkeyrobber

    Thread Starter New Member

    Aug 13, 2014
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    Hi,

    I've been playing around with some rudimentary circuit building and I am trying to prove the impact of a voltage divider.

    I understand the theory, I want to see the evidence on a multimeter.

    I'm not sure if I'm wiring the circuit, my multimeter is a dud or I am using the multimeter incorrectly - my money is on the latter.

    The multimeter is an DT-830B

    My circuit consists of a 3v battery and 2 10kΩ resistors in series.

    3v * ( 10kΩ / (10kΩ+10kΩ) = 1.5amps

    On the multimeter, I get a reading of 3.24 on the DCV setting, I expect to see a reading of 1.6 on the DCA setting. Every time, the reading settles on 0.

    I am a programmer - I've been coding for a living for years. I've been exploring circuit building for about a week, since going through a spark fun / arduino for beginners kit. I am very new to this side of the court.

    Any help would be greatly appreciated.

    Thanks in advance.

    Rob
     
  2. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    Ask yourself: does this answer make sense? Not only the absolute value, but the *units*.
     
  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    That makes no sense. You got ohms on top and ohms on the bottom, they cancel each other, what is left is X volts times some sort of number, volts times number is equal to volts. Where are you getting amperes?
     
  4. to3metalcan

    Member

    Jul 20, 2014
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    I think you've just got Ohm's law turned around. :) Current = Voltage / Resistance. And current is the same throughout a series circuit...it's just going to be your voltage divided by the total resistance of the divider.

    Lastly, I suspect that maybe nobody told you the amps function works differently from the voltage function...you have to interrupt the circuit with the ammeter, you can't just put the probes in parallel...in fact if you do so, you're shorting the component (the meter has near-zero resistance in amp mode!)
     
    Last edited: Aug 14, 2014
  5. donkeyrobber

    Thread Starter New Member

    Aug 13, 2014
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    I think I got that a bit mixed up.

    As I say, I am very new to this. I'm mixing up current divider and voltage divider circuits

    The current remains static in a voltage divider as to3metalcan says.

    I think the calculation should be more like

    3v * ( 10kΩ / (10kΩ+10kΩ) = 1.5v (correct units - thanks joeyd999)

    or if you prefer

    1.5v = ( 10kΩ / (10kΩ+10kΩ) * 3v

    [​IMG]


    Is this correct? If so, how can I prove it in practice?

    If it is wrong, could someone please tell me what the calculation should look like?
     
  6. to3metalcan

    Member

    Jul 20, 2014
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    That looks exactly right, and you can check it in a practical circuit by checking the DC voltage across either of the 10K resistors...one probe on each end of the resistor. Just don't switch over to the current function! ;)
     
  7. donkeyrobber

    Thread Starter New Member

    Aug 13, 2014
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    Great. Thanks for that!

    So I've got the probes either end of the series of resistors and getting a reading of 3v.

    Are you saying I should interrupt the series of resistors and put a probe between them?
     
  8. samuel.whiskers

    Member

    Mar 17, 2014
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    No, don't interrupt them - just put a probe at each end of each resistor in turn - there should be 1.5V across each one, and 3V across both together (both together is really just putting the probes on the battery terminals)....
     
  9. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Voltage is a phenomena that exits between two points, such as the two ends of a resistor. To measure it with a meter you just need to place your probes at the points.

    Current is a phenomena that exits within a conductor. To measure it with a meter you need to break the circuit to insert your meter probes.

    In either case the meter *will* change the circuit, which changes the reading, but for most things the change is insignificant.
     
    Metalmann likes this.
  10. donkeyrobber

    Thread Starter New Member

    Aug 13, 2014
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    Thanks one and all for your input.

    I have now see the proof that there are half the volts over each of the resistors, which is what I was looking for. :)

    I attempted to test the current in the same circuit, by adding the probe in series between the 3v and the 2 resistors. I wound the dial on the meter to DCA (20m range), hoping to see a reading of 0.15amps. I got 0. I tried the other ranges too. All 0.

    If I switch the probe for a jumper wire, I get a complete circuit. With the probe, I get a cold hard 0.

    I may have bust the probe with my experimenting last night, in which case I'll pick up another one tomorrow. Or am I missing something else?

    Thanks again for the replies. It's great to be learning! I'm at the base of an epic curve. I appreciate you guys helping me out on the first steps :)
     
  11. to3metalcan

    Member

    Jul 20, 2014
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    If you did short the circuit, you MIGHT have blown the internal fuse on your meter...it doesn't seem very likely with a 3V battery, though. Does your meter by any chance need the probe plugged into a different jack in order to measure current? Can you give us a picture (or at least a diagram) of what you're doing?
     
  12. donkeyrobber

    Thread Starter New Member

    Aug 13, 2014
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    It was the fuse!

    I replaced it and have been able to measure the current in my circuits.

    I am now able to calculate some expectations based on ohms law and verify those expectations using the probe!

    Thanks again! :D
     
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