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Finding Current in a Circuit (to get current going through an amp meter)

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data

    http://dl.getdropbox.com/u/119186/Picture%20122.png [Broken]

    2. Relevant equations

    Voltage Drop = -I*R
    I1=I2+I3

    Reqiv= 1/R1 + 1/R2 etc.

    3. The attempt at a solution

    I really have no idea how to approach this. A1 and A2 won't have the same current will they?

    I tried getting an equivalent resistor to get A1, but that didn't work:
    1/Req = 1/R1+1/R2 = 1/2.2 + 1/2.2 = 2.2/2 = 1.1=Req

    But still that can only get me the voltage drop and that hasn't worked...
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 1, 2009 #2
    Any ideas? I'm still stumped on this...nothing comes out right the way I'm trying this...
     
  4. Oct 2, 2009 #3
    Bump.
     
  5. Oct 3, 2009 #4

    Redbelly98

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    Homework Helper

    Redraw the circuit, but leave out the ammeters since they act as shorts as far as the circuit is concerned.

    Then look for parallel and/or series resistors, and come up with a single equivalent resistance for the entire circuit. That will help figure out the total current out of the battery.
     
  6. Oct 4, 2009 #5
    I got the entire Req as 4.6357 ohm, does that seem right? Then total current was 2.0493 A. Not sure if those are right or what to do from there...
     
  7. Oct 4, 2009 #6
    I did not get a total resistance of 4.6357 ohms. Can you show the work that you did to get to that point?

    Also, once you get the total current through the circuit, you can get the voltage drop across Req. So, will the voltage across Req be the same as the voltage across the 3 ohms? What about the 6 ohms? Why/Why Not, and show us where your thinking is.
     
  8. Oct 4, 2009 #7
    I found the problem! a) didn't notice that voltage is double what's given (see how the picture shows two batteries next to eachother? oops!)
    b) i got an Req for the parallel 2.2ohm resistors but apparently that's not necessary?
     
  9. Oct 4, 2009 #8
    a) It's one battery at 9.5V. We have multiple lines like that to represent multiple cells of a battery (one cell usually cannot do 9.5V)
    b) The 2.2ohm resistors are not in parallel. One way to think about parallel versus series is as follows. Every single straight line is going to represent a node. If two resistors are connected to the same two nodes, then they are in parallel. For instance, an R, 3, and 6 share the same two nodes, thus they are parallel. Let's collapse those three resistors down to an Requivalent at this moment. Are the resulting (equivalent) resistors in series/parallel and why?
     
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