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Circuits problem

  1. Oct 27, 2005 #1
    I have this circuits question that i've tried to do, but i'm not sure about my answers and i'm sure i'm going wrong somewhere...

    The 4.00 V cell in the circuits below has zero internal resistance. An accurately calibrated voltmeter across XY reads 1.50 V

    [​IMG]

    a) Calculate the resistance of the voltmeter.

    i did this :

    Vyz = 1.5 V
    therefore Vxy=4.00-1.50=2.50V

    Ixy=Vxy/Rxy
    =2.5/60
    =0.041666666...
    therefore Is=0.041666... (supply current)

    I(40 ohm resistor)=V(40 ohm resistor)/R(40 ohm resistor)
    =1.5/40
    =0.0375

    therefore Ivoltmeter=0.041666...-0.0375
    =0.00416666...

    1/Req=1/40 + 1/R(voltmeter)
    =(Rv+40)/40Rv

    Req=40Rv/(Rv=40)

    Vs=Is x R(total)

    R(total)=60+ ( 40Rv/(Rv+40) )
    4.00=0.04166666... x 60+ ( 40Rv/(Rv+40) )

    96=60+ ( 40Rv/(Rv+40) )
    36=40Rv/(Rv+40)
    36Rv+1440=40Rv
    4Rv=1440
    Rv=360

    Voltmeter resistance=360 Ohms

    This seem way too low for a decent voltmeter, so i think it's wrong, but i can't see how!!


    and for part b): Calculate the voltmeter reading when it is connected across Y'Z'

    I did this:

    1/Req=1/400 + 1/360
    =19/3600
    Req=189.473...
    R(total)=189.473...+600
    =789.473...
    Is=Vs/R(total)
    =0.0050666...
    V(600 ohm resistor)=Is x R(600 ohm resistor)
    =3.04V
    V(400 ohm resistor)=Vs - V(600 ohm resistor)
    =o.96

    Voltmeter reads 0.960V

    this seems so wrong!!

    i have no idea!!

    please help
     
  2. jcsd
  3. Oct 28, 2005 #2
    had a thought: in the last step of b) do i need to take into account the voltage across the voltmeter?? and how would i work it out?

    still got no idea on the first bit though...
     
  4. Oct 28, 2005 #3

    Ouabache

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    Homework Helper

    The wording in the question and the indication of the voltmeter in your drawing are different.
    If the wording of the question is correct, you may want to recheck how you solved this.
    Checking your analysis on the basis of having voltmeter as labelled, I agree with your part (a) answer.
    Part (b) also looks good. You drop 3.04v across the 600 ohm resistor leaving 0.96V across both the 400 ohm resistor and the voltmeter.
    huh? :confused:
    The voltage across the voltmeter is the same as the voltage across its resistance. Now the purpose of a voltmeter is to just measure
    the voltage and not have any affect on the circuit itself. What does this problem suggest regarding artifacts introduced by a measuring instument?
     
  5. Oct 29, 2005 #4
    i think it's my diagram that's wrong, i must have put the voltmeter across the wrong resistor!! silly me.

    i'll go and work it out again. :zzz: won't that be fun.
     
  6. Oct 29, 2005 #5
    actually, having just reread the question... it did say the voltmeter across YZ. not XY.

    so you think it looks ok?? i just wondered, because 360 Ohms seems too low a resistance for a good voltmeter, seeing as the ones we have even at college are in megaOhms.
     
  7. Oct 29, 2005 #6

    Ouabache

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    Science Advisor
    Homework Helper

    I think that the authors of that problem are trying to make a point. If you measure a load that is close to the same impedance as the voltmeter's input impedance, it is going to have a large impact (error) on your measurement.

    In other words, you can't just believe whatever your instrument tells you. You have to be smart enough to look at its characteristics and make your own decision as to whether reading is valid or not.
     
    Last edited: Oct 29, 2005
  8. Oct 30, 2005 #7
    thanks for all the help, i was really stuck on that one.
    :approve:
     
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