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D'Arsonval current meter

  1. Oct 29, 2011 #1
    The circuit is below. This is from a quiz from earlier this semester. The equivalent circuit for the meter reduces to the D'Arsonval movement in parallel with a resistor who's value is R/2. I know that no matter what current you want to measure, R must be greater than or equal to 18Ω so that the movement will read 90% of the current or better.

    The equivalent resistance of the meter will never be higher than 1Ω and will not affect the current i0 by more than approximately 1.13%. I was told I was wrong and received only 55/100 on this quiz.


    I would like some feedback if anyone has the time or the inclination.
     

    Attached Files:

  2. jcsd
  3. Oct 29, 2011 #2

    gneill

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    You'll have to post your attempt at a solution before we can comment.
     
  4. Oct 29, 2011 #3

    NascentOxygen

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    Pardon my ignorance, but is there a resistor missing from the top schematic?
     
  5. Oct 29, 2011 #4

    gneill

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    Suppose there was a resistor where (I think) you think there should be one. In what way would it affect circuit operation? :wink:
     
  6. Oct 29, 2011 #5

    NascentOxygen

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    The circuit would be better balanced. :smile:

    And it would quieten all these warnings I'm getting from SPICE. :smile:
     
  7. Oct 29, 2011 #6

    gneill

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    Ha! :rofl:
     
  8. Oct 30, 2011 #7
    The attempt is not really relevant. This is from a past quiz; it is not a current assignment. I'll try to explain myself again. The meter circuit can be reduced to an equivalent circuit that is the movement with a resistor in parallel. This resistor has a value of R/2. If you want this extended range meter to measure at least 90% of the current through a circuit accurately, R must be 18Ω or greater. The ratio of current then passing through the meter is a function of the resistances. I'm certain this is evident.

    When I ran this circuit through a simulator and put the meter in series with i0, it did not matter what the value of i0 was. As long as R was big enough, the current could be measured with better than 90% accuracy. My instructor claimed that the value of i0 was relevant, though I believe he is in error. As it turns out (I have yet to do the math) R actually has to be larger than 18Ω. If it is exactly 18Ω, I calculated an error of about 10.9%, about 0.9% to high. This additional error is due to the meter's equivalent resistance in series with the 80 Ohm resistor. I felt I should have received a higher grade on the quiz, as I was the only one who came close to calculating this value.

    I don't want anyone to solve the circuit but I was rather looking for some feedback that my reasoning was sound. I intend to argue my points to my instructor in person. Thanks for any help you would be willing to lend.
     
    Last edited: Oct 30, 2011
  9. Oct 30, 2011 #8
    No. They design these "circuits" with the intent to trick you into wasting time analyzing it. They have even gone as far as stating as much. You see, my ECE dept. is more concerned with tricks than knowing if we can actually solve circuits. It's a major point of contention with the faculty at my university, not to mention the students.
     
  10. Oct 30, 2011 #9

    AlephZero

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    You woukln't get any "warnings" if you built this from real components. "Knowing how to get a SPICE model of it to run" is a different question from "understanding how it works".

    I would disagree with that. They are probably trying to teach you to THINK, not just to pass exams.

    Forget about the re-drawing of your circuit you make in your OP. Start by thinking how many of the resistors will always have zero current in them, because the potential at both ends will always be the same.
     
  11. Oct 30, 2011 #10
    Have you redrawn the circuit? I know, without incertitude, the circuit reduces to an equivialent circuit that has just the 1Ω meter element and a single resistor of value R/2. I don't care about the six resistors on the end because I know they don't pass current. I am merely asking for someone to assure me I am analyzing this problem correctly. Furthermore, regardless of what they are teaching at my school, my instructor is often wrong, due to his lack of preparation and seeming apathy for teaching. I don't intend to receive half credit for something I know I did properly. Maybe you guys are too used to people asking for you to do their homework for them? :tongue:

    I'm beginning to suspect I should have posted this somewhere else....
     
  12. Oct 30, 2011 #11

    gneill

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    From what you've posted so far it's not clear exactly what the question is. You've presented some conclusions that you came to during analysis, but not how you arrived at those numbers, or how relevant they are to responding to the the quiz question. The actual phrasing of the quiz question, in its entirety, would be helpful.

    I take it from your last post that the meter movement has an internal resistance of 1 Ohm. Does it have a maximum current rating for full scale deflection that you must abide by?
     
  13. Oct 30, 2011 #12
    If you click on the picture with the circuits, all the information is there. The meter circuit, the circuit with i0 and the instructions. I would gladly explain, but it will be far easier for you just to look at the attached picture. If you can't read some of the values, let me know.
     
  14. Oct 30, 2011 #13

    gneill

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    So, apparently the actual current that passes through the meter movement is irrelevant. How very unrealistic! I would have thought that a reading near full-scale would be required for good accuracy.

    One can reduce the circuit under test to a Thevenin equivalent voltage source and resistance driving the 80Ω "load" resistance. Then the effects of inserting the meter will be obvious:

    attachment.php?attachmentid=40510&stc=1&d=1320007451.gif

    As you surmised, the 1Ω meter movement resistance will make the paralleled resistor moot; the 80Ω resistance swamps the effect. Without the meter in place the current is 240 mA. With the full 1Ω resistance inserted it's 237.2 mA. Magnitude of the percent change is (240 - 237.2)/240 or 1.2%, well within the desired 10%.
     

    Attached Files:

  15. Oct 30, 2011 #14
    Exactly as I concluded. To say again, these problems are entirely unrealistic and it can be confusing since I have a background in electronics as a technician. I find this class frustrating at best. My instructor also tried to argue that a current source in series with a capacitor would not pass current through it. You can imagine my chagrin when I saw my score on this quiz. He obviously does not fully prepare himself for this class before he lectures us. *sigh Thank you for your help.

    One more thing, as I mentioned, if you set R = 18Ω, you get an error of about 10.9%. So, I would imagine the value would have to be just slightly higher than 18Ω to pull the reading into the desired window. I didn't anticipate this on the quiz, but at least I came up with a value that was reasonably close. Also, you can solve for this value of R without ever knowing what i0 is. I shall be speaking with my instructor tomorrow...
     
  16. Oct 31, 2011 #15

    NascentOxygen

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    It has taken me a while to catch on to what is going on here, partly because the question was not well explained at the start, and making this to be a mindreading exercise in more ways than one.

    Let's see if I have the picture as you saw it...

    A current meter was available for a task, but its FSD was slightly less than the application required, so the meter's range was extended by about 10% by adding a crazy circuit around it, equivalent to a shunt resistor of 18 ohms in parallel to the meter movement's resistance of 1 ohm. Presumably the face of the meter is now rescaled so FSD reads 10% higher?

    If we determine by how much this metering arrangement will upset the current in the 80 ohm resistor in Fig 1, thanks to gneill we know that it reduces the current by only about 1.2%. The big problem I have here is that D'Arsonval movements can't carry that many mA.

    Now, if the application could tolerate an upset of 10% when the meter was inserted, then it seems the design has been tighter than necessary. Or maybe it's not the way it was intended to be solved?

    Let's start over, looking at it differently.

    You have a bare meter movement (D'Arsonval meter). They are very sensitive devices, requiring less than a mA for FSD, so shunt resistors are mandatory for measuring fractions of amps. This would make the metering arrangement always equivalent to less than 1 ohm, no matter what value of shunt resistor (typically around 0.1 ohm) is added. If FSD is set to too large a value, for example, 10 A, it will be difficult to accurately take a reading around 0.25 A. So you'd choose FSD to be no more than 1.25 amp so your 0.25 A is at least 20% along the scale. A meter's accuracy is often given as % of FSD, so for 10% accuracy at 0.25A, it would equate to 2% FSD accuracy at 1.25A FSD.

    All that remains is to determine R that will turn the meter movement into an ammeter with 1.25A FSD. :smile: Alas, we are lacking a vital piece of information here.

    Have you discussed this with other students? Did others in the class not ask for extra details during the test?
     
  17. Nov 3, 2011 #16

    NascentOxygen

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    We are waiting to hear what he says....
     
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