Finding voltage at a point

Discussion in 'Homework Help' started by Cue Cue, Oct 1, 2008.

  1. Cue Cue

    Thread Starter New Member

    Sep 25, 2008
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    0
    I have been working on this problem for the past two days and have not been able to figure it out.

    The question is:
    "The voltage at point A (Va) is to be measured with the Fluke 45. What will the meter read and what is the loading error?"

    The answer is suppose to be:
    Fluke will read 59.88 V and the loading error is 9.2%

    This question is a part of my lab on Meter loading effects and this is the only part of the lab that i cannot complete

    some quick info on the Fluke 45 DMM is that as an ammeter, the Fluke 45 represents a load resistance of 10 ohm. As a dc voltmeter, it represents a load resistance of 10 M ohm

    The circuit that belongs with the question should be attached to this post.

    I think the problem I am having is understanding the circuit. I am not really sure where the meter would go when trying to measure voltage on this circuit. I have attached a pdf of some other work ive done. sorry thats only the bit of work ive got scanned. Ive tried this question over and over varying the way i do it and the closest answer ive gotten to was around 57 volts

    also the circuit in the drawing is what the i assumed the circuit in the question would look like am i wrong to think this?

    "PDFonline.pdf" is the circuit drawing the question gives
    "bit of work of done.pdf" was an example of the one of many ways i attempted to solve this

    please help as soon as possible
     
  2. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    Hi
    I did the calculations and it does come out to 59.88 volts.

    Have you learned how to calculate the thevenin voltage and resistance of a circuit?

    What you do is unhook the meter resistance and solve for the equivalent voltage at point A. Then you solve for the equivalent thevinin resistance,

    Then with this new voltage and resistance you plug in the voltmeter resistance and the series resistance and then solve for the voltage drop actoss the voltmeter 10M.

    Or

    I noticed in your second diagram you didn't put the 10M res. in.

    Take it slow and start from the begining, with the voltmeter in the circuit it acts as a 10M res. in series with the 560k res. this series comb. is in parrallel with the 1M res. you need to draw your diagram with the 10M res. in series with the 560k res. then go through your calculations to solve for the voltage drop across the 10M res. because that is point A. to ground.
     
    Last edited: Oct 1, 2008
  3. Cue Cue

    Thread Starter New Member

    Sep 25, 2008
    4
    0
    ok well i tried again with adding the 560k and 10 M resistors in series. I then put those two in parallel with the 1 M resistor and then added that to the 820k ohm resistor to get the total resistance which i got was 1733494.8 ohms

    im still confused am i suppose to divide the 10 M from the total resistance then multiply that by the total voltage? I am getting a really high number

    Im probably missing something really obvious..but i cant think to straight i have a DC test tomorrow and I have this lab report due and im trying to get it done so i dont let my lab partner down since its my turn to do the calculations on the lab so im quite stressed out
    i will continue to work on the problem if you can provide any more help that would be really helpful
     
  4. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    Alright try this. This is how you solve for an equivalent voltage and resistance.

    step 1. with your volt meter out of circuit you have a equivalent voltage of
    Vin. (120) X 1M divided by (1M +820K) that is your equivalent voltage seen at point A with no load. Now you need to find the equivalent resistsnce of the NO load circuit.
    to do that you take the value of the parrallell combination of the 820k res. and 1M.

    now redraw your diagram with this new equivalent voltage in series with this new resistance, with this drawn up you now hook up your 10M res. and 560k series res. as the load to this equivalent circuit, then just use voltage divider equation to solve for the volt. drop across the 10M res.

    I'll see if I can draw this up for you to study it.
     
  5. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    Here is a diagram of solving for euivalent circuit.
     
    • ETH.JPG
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  6. Cue Cue

    Thread Starter New Member

    Sep 25, 2008
    4
    0
    ok i finally figured it out thank you sooo much that takes such a load off me.
    I can finally just spend the rest of the night/morning studying for my test tomorrow. thanks again i really appreciate it.
     
  7. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    Your welcome.

    Glad to be of help...
     
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