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Homework Help: Dc Circuits - Voltmeter/ammeter

  1. Mar 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Hey, How is everyone? I am working on a lab dealing with DC circuits. I am a bit confused about voltmeters and ammeters. We used a carbon composition resistor which obeys Ohm's law and therefore is a linear device. My Paint drawing aren't too pretty but I hope they convey the problem better.

    Problem 1: The following circuit was setup:


    Should the current displayed for both Ammeters be the same?

    My Answer 1: Yes, because current is not consumed in a circuit. It's value remain the same throughout the circuit. It is only the electrical energy that is consumed or transformed in the circuit. In the circuit above, the original current (i.e the resistor) has high electrical energy relative to the energy of the current that flows out of the resistor. The resistor converts the electrical energy to heat, dissipating it into the ambient air.

    Problem 2:A multimeter set to voltmeter function was used in a position where an ammeter should have been as shown in the circuit below. Is there anything useful displayed on the voltmeter?


    My Answer 2:Since a voltmeter measures the potential difference across two points, the difference between the power supply and the resistor should equal zero because the current coming out of the power supply has the same electrical energy (ignoring resistance of the wire) as the current going into the resistor. Also, the internal resistance of a voltmeter is high.

    Problem 3:A short circuit occured in the electrica circuit below. A multimeter set to ammeter function was used instead of a voltmeter (which should have been used). What happened?


    My answer 3: What i understand for this question isthat an ammeter is at a position where a voltmeter should have been used. A voltmeter is always hooked up parallel in a circuit to measure the potential difference between two points and it has a large internal resistance so that the correct voltage can be measured. An ammeter is hooked up in series so that all the current goes through it and it has a small internal resistance so that the maximum current can be measured. In the above case, you have the ammeter in parallel instead of series and the current is divided. Umm.. I know about this much but can't seem to answer it.

    Problem 4:The resistance (R) of a resistor in a circuit was measured using a voltmeter. It was known that the R of the resistor was a very large value, close to the internal R of the voltmeter. Would you get an accurate measure of R for the resistor?


    My answer 4: I have no clue for this one. It seems that if the resistor has a huge R value, then the potential difference measured by the voltmeter will be very huge as well. But why would the readings on the voltmeter be inaccurate?... dont know.
    Last edited: Mar 16, 2007
  2. jcsd
  3. Mar 16, 2007 #2
    I like your reasoning on all the problems except 4. Re 3, what happens is just what is described--you short the circuit so that unless the parallel resistor is very small, all current runs thru the ammeter. Eventually this reaches a finite value which reflects the internal resistance of the battery--in other words the current will not be infinite. Additionally this current will likly sag with time as the internal resistance increases due to heating.

    On 4, the potential difference should be accurate irrespective of the external resistance in parallel and equal to the voltage of the battery itself, ideally assuming an infinite input impedance (roughly resistance here in the steady state). I suspect the reading would become nosier, but not sure you should know that. Look here for some more help:

  4. Mar 17, 2007 #3
    umm sorry i still don't understand the answer to 3. Why would all the current run through the ammeter and reach a finite value. What do you mean by that? Isn't the current divided because the ammeter is in parallel (ie total current = current(resistor) + current(ammeter)) . For problem#4, my teacher's assistants told me that as the resistance of the resistor approaches the internal resistance value of the voltmeter, the readings will get get more and more inaccurate. I don't understand why.
  5. Mar 18, 2007 #4
    its just those 2 last ones that i am stuck on
  6. Mar 18, 2007 #5


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    That is correct, however, since the ammeter has a much lower resistance than the resistor the the majority of the current would pass through the ammeter as opposed the the resistor. Think, path of least resistance.
  7. Mar 18, 2007 #6
    ahh excellent so i seem to understand from what you said that the current "favors" the path with the lower resistance. But the thing is when you place an ammeter in series, all the current runs through it anyways (and it should if you want an accurate reading). Why would making it in parallel now, cause a short circuit, since the majority of the current is running through it anyway?

    Also, for problem 4, any ideas why a voltmeter would measure an inaccurate resistance for a resistor that has an extremely large resistance (close to the value of the internal resistance of the voltmeter itself)?
  8. Mar 18, 2007 #7


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    Think of two resistors in parallel. If one is really large and the other small, what is the equivalent resistance? If they are both about the same, then what do you get? Work it out to see. So seeing the result of that, what do you think having a resistance approaching the resistance of the voltmeter will do to the overall resistance of the circuit (and the measured value of voltage)?
  9. Mar 18, 2007 #8
    The more a device interferes with the circuit it is measuring, the less accurate it will be. Ammeters are designed to have very little resistance, so when connected in series there is no voltage drop. If you decide to connect an ammeter in parallel, it would be equivalent to creating an alternate path of wire only (no resistance). At a parallel junction, the current is split between 2 paths in proportion to the resistance of the paths. If a path has 1/3 of the total resistance, 2/3 of the current will flow through it. If a path has zero resistance, all the current will flow through it and none will flow through the other resistor.

    A voltmeter is the exact opposite, and is designed to have very high resistance, so when connected in parallel there is no current flowing through it. The idea is for all the current to flow through the other path. If the voltmeters resistance is close to the other resistors resistance, then the current will be split between the two.
  10. Mar 18, 2007 #9
    hmm ok i seem to be getting the concept... let me see if i can state mathematically what hage567 mentioned..

    2 resistors in parallel:
    --> 1/R = 1/R1 + 1/R2 where 1 and 2 denote the resistors
    --> I/V = I/V + I/V (same way of writing above eqn)

    so if resistor 1 is very large and resistor 2 other is small, all the current will go through the one that is very small. If they are both equal, the current will be split between the two. hmm... so like turd said, the whole idea of a voltmeter is to have a high resistance so that the current will flow through the other path BUT if i had a resistor that had a resistance close to that of a voltmeter, now i have two equally large resistances, and the current "doesn't know" which of the paths to take. Since they are equal, the current will just split and you won't get an accurate potential difference when measuring the resistor's resistance. That is the best answer I can give and it makes very good sense. SO PROBLEM 4 I UNDERSTAND THE CONCEPT NOW.

    PROBLEM 3:
    For an ammeter in series, all the current flows through it and there is no voltage drop (or a negligible one). There is only one path here. However, for an ammeter in parallel, you have 2 paths for the current, one going through the resistor and one through the ammeter. so, therefore now the current has branched off and you won't get an accurate reading for the current. However, what causes the short circuit? If an ammeter is in series, all the current is going through it. If an ammeter is in parallel, the current is branced off and you have less current going through it but still sufficient since its internal resistance is low. Is it because now that you have less current going through it, you have increased resistance (by Ohm's law), this increase in resistance is what causes the short circuit in the ammeter. umm.. does that make sense.
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