Voltage Meter Connection With Scales

Discussion in 'Homework Help' started by xz4chx, Sep 24, 2012.

  1. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    Okay so I have this problem Figure 1 is multi-range voltmeter with two scales. First scale is 50V and Second is 200V. So I solved for R1 and R2 that was simple. Then it says to connect it to Figure 2, Common to Common, 50 V to 50 V and 200V to 200 V. How should they connect?

    http://imageshack.us/photo/my-images/26/14876924.png/

    So if I connect them should it look like this?

    http://imageshack.us/photo/my-images/713/image201209240001.jpg/

    If that is so then how do you account in the bare wire through the 50V line when using calculations because every problem I have done has a component through it. The procedures of the problem are pretty simple, I'm just not understanding that one part
     
    Last edited: Sep 24, 2012
  2. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    I posted links because I could not figure out why my pictures were not loading onto the page
     
  3. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    I think those instructions are incorrect. Who drew and labeled figures 1 and 2, your or your instructor?
     
  4. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    The two figures that are printed are the teachers, the notebook drawing was mine.
     
  5. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Here's the problem ...

    In figure two, you have a 7 mA current generator feeding a series resistor. When you connect the meter to the circuit ... not the way it's drawn .... the meter receives 7 mA. The meter movement is 5 mA full scale. Can you say "damm, I might have just ruined up my meter."

    You converted that 5 mA meter to a volt meter in figure 1. You are attempting to measure a current.

    You got the multiplier for the 200V scale correct and the multiplier for the 50 V scale incorrect.

    Are you suppose to convert that current source to a voltage source and then measure it with the voltmeter ... identifying the percentage of the full scale you will see?
     
    Last edited: Sep 25, 2012
  6. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    I dont know. I'm very confused about problems like this. Our teacher didn't go over it too much in class and the textbooks do not have many expamples. I'm going to go after hours to professors office to get some advice.
     
  7. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    I looked at the circuit again. You can connect them as drawn and compute the current or voltage read by the meter. To see, you need to redraw the circuit to form a simple series parallel circuit.

    That circuit forms a "voltage source" for the 50V input.
     
  8. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    So what am i doing wrong on my drawing? I'm am still not understanding how they connect, because I am not getting a correct solution. With what I am drawing I keep getting that the 200V scale is reading 210V because of the 7ma current source. I was not able to ask my teacher as our class was cancelled on the day of my class.
     
  9. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    And i did correct my resistance, that was just a dumb mistake
     
  10. JoeJester

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    Apr 26, 2005
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    Here's what I did ... I'll leave it to you to calculate what the meter reads .... This is clearly an academic exercise as a voltmeter would have switch(s) to select the 200V or 50V scales and not have them connected concurrently.
     
    Last edited: Oct 1, 2012
  11. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    Okay that makes more sense, I was trying to manipulate the circuit instead of just connecting them directly. The examples we have done in class, was not connecting all the pieces but just reading from either the 50V or 200V. I want to thank you for all your help
     
  12. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    So I just did this through a KVL, KCL and by putting the resistors in parallel. One quick question, is the AM2 just a reference or is that something that goes into the circuit? Usually you just connect the circuits together and measure the voltage from the common to the 200V connection. If I do this i get

    R8 + R6 = 10000 Ohms = Req1
    Then
    1/R5 + 1/Req1 = 2481 Ohms = Req2
    So Req2 + R7 = 32481Ohms

    And .007A * 32481 Ohms = 227.367 V which is not correct and I set up a KVL, KCL and got the same answer, so what am i doing wrong

    KVL, KCL
    -Vs + 210V + 3300i1 = 0
    -Vs + 210V + 10000i2 = 0
    .007 = i1+i2

    (i1 is current going toward R5, i2 is current going toward R6, and Vs is voltage of 7ma current source.)

    Vs = 227.367 V which is still stronger than the meter full scale reading ability.
     
  13. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    You forget one thing ... the meter is an Ampmeter ... not a volt meter.

    The two series resistance could convert it an voltmeter, but I digest.

    Look at the combined circuit and answer the question "how much current is flowing in the 5 mA full scale meter".

    Cross check it with Kirchoff's current law.
     
    Last edited: Oct 1, 2012
  14. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    So basically I found out the current through the meter with everything connected and since resistance is the same, I just took the ratios of the voltages and currents in order to find the new current. So I had the full scale meter voltage and current, and then found the current after all the circuits were connected and took the ratio to find the new voltage. I remembered my teacher talking about this process in class, so this made the problem much more simple.
    So I got it perfectly. Thank for all the help again.
     
  15. JoeJester

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    Apr 26, 2005
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    So... what did you get for the current flowing in the meter?
     
  16. xz4chx

    Thread Starter Member

    Sep 17, 2012
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    I used a ratio to get the new current of the meter when everything is connected. So 7ma is following through R7 and then I used current divider rule to get the current running through the meter.

    So 7 * (R5/R5+R6+R8) = 1.7368mA

    Then I did
    (1.7368/5 - (Full scale current through meter))*200V = 69.5V

    69.5 V was the correct answer
     
  17. JoeJester

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    Apr 26, 2005
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    Very nice work.

    I guess the instructor could have asked what would the voltage read on the 50 volt scale as well.....as it was also connected.

    My original error was thinking it was a normal voltmeter with switched inputs ... which is why I retracted and followed up. I should have known better since the homework section is always able to have academic exercises.

    It was nice working with you and thanks for participating.
     
  18. xz4chx

    Thread Starter Member

    Sep 17, 2012
    71
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    Thank you.
    I really appriciate all the help
     
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