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Determining current from ammeter

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data
    32_P50.jpg
    The ammeter reads 5.0A. Find I1, I2, and ε

    2. Relevant equations
    ΔV=IR

    3. The attempt at a solution
    I've never dealt with ammeters before so I'm very confused.
    To start, I tried breaking this into loops, and then used Kirchhoff's Law.
    I started with the left side, since I have the value of the battery. I know that ΣΔV = 0 in this loop. I also know that these two resistors are in series, so Req=3.0+2.0=5.0Ω
    Since I = V/R, I tried I = 9/5, which is incorrect

    I then decided to think about this with the junction law, and thought that because the current at A is 5.0A, that's what's entering the battery, thus that must also be what's exiting the battery, but this was again incorrect.

    I've gone through many different methods and have been unsuccessful in all, and am really just trying to figure out where to even start. Any help would be much appreciated!
     
  2. jcsd
  3. May 15, 2015 #2

    phinds

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    Forget the ammeter. The question really is just asking what the circuit looks like if the current through the 2ohm resistor is 5 amps.
     
  4. May 15, 2015 #3
    Okay. I'm still unsure of how to do this.
    I split the circuit into the loop on the left. Using the fact that the current through the 2ohm resistor is 5A, I found that the change in potential at this point is 10V. I know that the total change in potential must be 0, so the change in potential through the 3ohm resistor is must be -10V, thus the current would be 10/3. This is the only way I can think of to do this, but looking at it I feel that I'm not doing this correctly...
     
  5. May 15, 2015 #4

    phinds

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    I can't imagine how you get -10v across the 3 ohm resistor. What is the voltage on the left side of that resistor? What is the voltage on the right side of that resistor?
     
  6. May 15, 2015 #5
    Ohhhh oops, I forgot to add the voltage from the battery... This isn't intuitive to me at all. So Δ5 across the 2Ω resistor is 10V.
    9V - 10V = -1 V
    -1 V / 3 Ω = -1/3 A
    Thank you so much!
     
  7. May 15, 2015 #6

    gneill

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    Staff: Mentor

    Since you've noticed that indicated currents can be negative, and since no mention was made of the direction of the current that the ammeter reading represents, then for full marks you may have to consider the two cases: 1) 5 A is flowing downwards through the 2.0 Ω resistor, and 2) 5 A is flowing upwards through the 2.0 Ω resistor.
     
  8. May 16, 2015 #7

    phinds

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    No problem. Yeah, it's easy for those of us who have done a few of those problems to forget how confusing things can be at the very beginning when you are trying to take in everything at once.

    qneil brings up an interesting point. What does is the circuit doing if the 5 amps is flowing UP through the 2 ohm resistor? It's a good exercise.
     
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