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Homework Help: I don't get ammeters or voltameters!

  1. Oct 26, 2005 #1
    Hello everyone, so far i've missed every single problem, this book has a section on ammeters and voltameters that is like 1 paragraph long and just says, the ammeter measures current, the voltameter measures Potential difference. And this problem asks:
    n Fig. 28-40, the ammeter and voltmeter resistances are 2.00 and 268 , respectively. Take E= 12.0 V for the ideal battery and R0 = 119 .
    Here is my work: http://img201.imageshack.us/img201/2220/pink8iw.jpg [Broken]

    Figure 28-40a Figure 28-40b
    Here is the first figure, 28-40a: http://www.webassign.net/hrw/28-40a.gif

    (a) If R = 93.0 what will the meters read for the two different connections (Figs. 28-40a and b)?
    Figure A:

    Current wrong check mark A
    Potential Difference V
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 26, 2005 #2

    Chi Meson

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    The current through R is not the same as the current through Ro. Are you supposed to use the loop rule to solve this? I'd think it'd be easier to do simple parallel/series analysis:

    Resistor R is in paralle with the voltmeter. This parallel is in series with the ammeter and Ro. Find the total current and then the voltage drop across the parallel.
  4. Oct 26, 2005 #3
    thanks!! he never said how i have to solve it...So u said to find the total current, then the voltage drop, Do i use the formula for adding up resistors in paralell? like I know the same potienal differences across all resistors in parellel, so 1/Req = 1/Ro + 1/Rvoltmeter;
    then find Req = Ro + Rammeter ?
  5. Oct 27, 2005 #4

    Chi Meson

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    Sorry for taking a while, the internet was down at school today. Yes the "inverse sum" method is how you find the parallel equivalent resistance (call that Rp). The the total resistance of the circuit is Rp + Rammeter + Ro. From this and the emf, you find the total current. This current times Rp is the voltage drop across both "R" and the voltmeter.
  6. Oct 27, 2005 #5
    Hey no problem thanks alot Chi, that wokred great :)
  7. May 15, 2010 #6
    whoa this site is awesome =D
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