calculating indication of an ideal ammeter

Discussion in 'Homework Help' started by kawafis44, Mar 13, 2008.

  1. kawafis44

    Thread Starter New Member

    Mar 13, 2008
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    that's the exercise [​IMG]

    and my solution [​IMG]

    calculations for A,B,C (1.1)
    -I_1 +I_3 + I_5 = 0
    -I_3 + I_4 - I_6 = 0
    I_2 -I_4 -I_6 = 0

    calculations for I, II, III (2.1)
    -U_1 -U_3 +0 = 0
    -U_2 -U_4 -0 = 0
    U_3 + U_4 + 25 = 0
    because U_6=0, E_5=25V

    i use resistors R_1,R_2,R_3,R_4 in this formula U_i=R_iI_i
    and i have got (2.2)
    -2I_1 -8I_3 = 0
    -1I_2 -9I_4 = 0
    8I_3 + 9I_4 + 25 = 0

    i try to calculate from 1.1 and 1.2 this current I_6 (i thinks this is what i need to find), but i cannot obtain the result. by the way calculations are too complicated for an exercise like this

    greetings!
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    The ideal ammeter looks like a short circuit (or another length of wire). That means the current will only flow in two of the resistors. Ohm's law will let you figure the current easily.
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Come again:confused:
     
  4. Ron H

    AAC Fanatic!

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    Here's a big hint. I did it using Thevenin's theorem.
     
  5. beenthere

    Retired Moderator

    Apr 20, 2004
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    Probably too rough-and-ready of me, but I looked at the ideal ammeter as a short. That made the current path into the 8 ohm resistor, through the ammeter, and out the 1 ohm resistor. Solve for the pairs in parallel to get the voltage drops, and there is the current through those resistors (the 8 & 1 ohm) and therefore the ammeter.
     
  6. Ron H

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    I'm not following you. Does that account for the current through the 2 ohm and 9 ohm resistors?
     
  7. beenthere

    Retired Moderator

    Apr 20, 2004
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    No. And on second look, my visualization doesn't look quite so correct. I saw the 8 ohm and 1 ohm as the current path through the ammeter.
     
  8. hgmjr

    Moderator

    Jan 28, 2005
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    There are a number of ways to solve your problem but RonH's suggestion of using Thevenin's Theorem appears to represent a straightfoward solution.

    One thing RonH, did you purposefully omit tying the two grounds together in your diagram.

    I have altered you redrawn version of the schematic to show what I am referring to.

    hgmjr
     
  9. Ron H

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    Oops!:(
    I forgot to draw that in. Good catch!
    Obviously, no current will flow without a return path.
     
  10. hgmjr

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    Jan 28, 2005
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    I suspect that you had envisioned ground symbols drawn on each of the loops. I had to look at it a couple of time before I noticed there were no ground symbols. After you have looked at as many schematics as you and I have over the years, you tend to overlook a missing explicit connection and assume that they are connected together implicitly.

    hgmjr
     
  11. Ron H

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    Yeah, I have to confess that after I did the math, I checked my results with LTspice. LTspice won't let you run a sim without a GND. Before I posted, I took all of them out because the OP's schematic didn't have one, and forgot to put in the wire to complete the circuit.
    I didn't actually simulate the intermediate step, but the copy and paste still had GND, which I removed.,
     
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