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Electricity and measurement.

  1. Feb 4, 2014 #1
    Hi.

    I have a problem understanding a question for my electronic engineering course. We have a subject that's directly translated to 'the knowledge of electricity and measurement techniques'.

    I do hope this is the correct forum to ask for help at.

    1. The problem statement, all variables and given/known data

    Given the following circuit:

    See attachment.

    Construct a voltmeter with max impact/effect at 10V, when you have a moving coil with the following data: Ifull scale = 1mA, Rm = 2k ohm. What will the voltmeter show when it's applied to the 10k ohm resistance (as shown on the figure), and what will the error be in percentage of ideal voltage because of the load effect?
    2. Relevant equations

    The sensitivity of the instrument is S = 1/I -> S = Rm/Ufullscale

    3. The attempt at a solution

    I don't have a specific attempt at a solution as I'm confused, but this is what I've done.

    The current in the circuit before the voltmeter is applied is I = U/R = 100V/105kohm = 0.952mA.

    With the voltmeter applied, it will create a 2k ohm resistance in parallel with the 10k resitance making the current in the circuit: I = 100V/55kohm+40kohm+10kohm||2kohm = 1.035mA.

    The current that runs through the 10k resistance will be:

    10k resistance is 5 times larger than the 2k resitance ->5 times more current will flow through it.

    1.035mA/6 = 0.1725mA * 5 = 0.6375mA

    The voltmeter surely will show: U = R * I -> 10k ohm * 0.6375mA = 6.375V?

    The ideal voltage is 0.952mA * 10k ohm = 9.52V

    Error will be 67%.

    Now, here's the tricky part. The answer to this exercise is R = 8kohm and error is 47.5%.

    I'm totally confused! How can a voltmeter show ohm? Please help me with any input you might have, I've been stuck on this problem for three days now.

    And do let me know if I need to clearify anything. Thanks in advance!
     

    Attached Files:

    Last edited: Feb 4, 2014
  2. jcsd
  3. Feb 4, 2014 #2

    gneill

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    Staff: Mentor

    First you want to design your voltmeter. Apparently it should have a full scale deflection when connected across 10V, presuming that's what "max impact/effect at 10V" means. Use the given information about the meter coil to figure out what you need to add to it in order to achieve this.
     
  4. Feb 4, 2014 #3
    Thanks a lot for showing interest!

    I'm afraid I'm still not able to see what I'm really looking after.

    Design your voltmeter. The voltmeter is there? Its internal resistance is 2k ohm? Is there anything more to it?

    I'm pretty sure your assumption is correct.

    Use the given information about the meter coil to figure out what you need to add to it in order to achieve this.

    The given information about the meter coil is clear, but what is it that I'm supposed to achieve? The part that completely confuses me is that I'm supposed to get resistance - ohm - from a voltmeter. I've been twisting and turning the formulas and doing all kinds of calculations, but 8k ohm hasn't even once turned up on my paper, nor calculator.
     
  5. Feb 4, 2014 #4

    gneill

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    Staff: Mentor

    You've been given information about the coil ("moving coil"). Not about the voltmeter. You need to add some components to make it a practical voltmeter.

    Have you noticed that real voltmeters have a selector switch to choose the voltage range? The meter movement (coil) is not changed, it always requires the same amount of current to achieve a full-scale deflection. That means the switch is selecting different resistor networks to achieve that full-scale current when the voltmeter is placed across a voltage corresponding to its selected range.

    Now. Consider the coil that you've been supplied with. How much current would it draw if you just stuck 10V across it?
     
  6. Feb 4, 2014 #5
    These components that I need to add, are those shunt resistors? That's the only thing that I have in my noted that makes any sense.

    Yes, it's selecting between shunt resistors?

    I = U/R -> 10v/2kohm -> 5mA? I still can't see how to solve this problem. I really don't.

    I'm terribly sorry. I clearly must've missed something, somwhere. I don't even know the proper terms to describe the components, so it's really hard to use google to find information as well.

    Please stay with me, I'm doing my best.
     
  7. Feb 4, 2014 #6

    gneill

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    Staff: Mentor

    If you were add a shunt resistor (a parallel resistor), would that affect the voltage across the coil?

    Is 5mA appropriate for the coil? What's the maximum allowed full scale current for the coil?
    If you wanted to reduce the current flowing, how might you do it?
     
  8. Feb 4, 2014 #7
    I assume the answer is no, as you add parallel, but it will divide the current.

    1mA is maximum for full scale. To reduce it, I add a resistance in series?
     
  9. Feb 4, 2014 #8

    gneill

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    Staff: Mentor

    More precisely, adding a parallel resistor will increase the total current drawn from the source, but will leave the same amount of current flowing through the coil --- parallel resistors share the same potential difference.

    Correct. What total resistance is required to allow just 1mA to flow?
     
  10. Feb 4, 2014 #9
    10k. So if I add an 8k resistance in series with the 2k, I have my answer?
     
  11. Feb 4, 2014 #10

    gneill

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    Staff: Mentor

    Well, you'll have "constructed" your voltmeter. The 8k value is correct. :approve:

    Now you need to connect that meter to the circuit as shown your figure and answer the remaining questions.
     
  12. Feb 4, 2014 #11
    I just had a voice in my head that said 'Ah!'. Thank you! I must've missed this, but I'm still wondering why there's not a single word about it in our book or lecture notes, but oh well. Thanks for staying with me! :)

    Although, I'm not quite finished. The voltmeter will get full scale measurement at 10V. The current that runs through it is 0.5mA through 10k resistors = 5V. The actual voltage is 9.5V. Error = 52.6%. This is slightly off the actual answer that's provided with this problem. Is just a matter of decimals, or am I going about it the wrong way?
     
  13. Feb 4, 2014 #12

    gneill

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    Staff: Mentor

    How are you calculating your percent error?
     
  14. Feb 4, 2014 #13
    (5/9.5) * 100. It might be (1-(5/9.5)) * 100? Haven't done percentage in a while :blushing:

    That face if for embarrassment. Heh.
     
  15. Feb 4, 2014 #14

    gneill

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    Yeah, the second one is better. If you have a "true" value Vo and a measured value Vm, then the percent error is

    ##Error\% = \frac{V_o - V_m}{V_o} \times 100##

    Basically the difference from true divided by true, multiplied by 100.
     
  16. Feb 4, 2014 #15
    Got it.

    Thank you so much for taking your time with this. I've been using my noted (which clearly misses something) and tried to calculate every single formula known to man to try and solve this problem for three days (not even exaggerating).

    I love this forum.
     
  17. Feb 4, 2014 #16

    gneill

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    Staff: Mentor

    Glad I could help. Good luck in your studies.
     
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