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Simplifying a circuit to establish the intesities

  1. Mar 10, 2013 #1
    So I'm to calculate the current intensities in the places I put the ammeters in (yeah, I reckon one of them is turned around :)):

    http://i.stack.imgur.com/PAmmD.jpg

    And while I can calculate such things for series or parallel circuits, I was taught I should first try to simplify the circuit and do it with Ohm's law and supplementary resistance. So I did something like that (which is my only idea):

    http://i.stack.imgur.com/E4svU.jpg

    And, regretfully, the results in the simulator are different. Then, how can I evaluate the 3 Ohm resistor (the one being "vertical" in the first picture) to somehow simplify the circuit and be able to calculate it? Could you please help? :)
     
  2. jcsd
  3. Mar 10, 2013 #2
    There are no resistors in series in the first circuit you posted. Notice that there is a branch between any two resistors.

    Also, there are no two resistors in parallel. The technique of simplifying series and parallel resistors will not work for this circuit.

    Instead, there are other methods, which are used to create a system of linear equations that describes the circuit.

    In the node-voltage method, that fact that the current going into a node (a place on the circuit) must be equal to the current going out of a node is used.

    In the mesh-current method, the fact that voltages in a loop must add to zero is used. The circuit is divided up into loops, and each gets its own current. For parts of the circuit which are in multiple loops, the total current (which would cause a voltage on any resistors there) would be the sum of the currents for all the loops which go through that part of the circuit.

    You might be able to find some decent learning material by searching for one of these methods.
     
  4. Mar 10, 2013 #3

    gneill

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    Staff: Mentor

    Your second circuit is not equivalent to the original.

    Note that the 2,3, and 4 Ohm resistors form a Δ configuration. Similarly, the 3,4, and 6 Ohm resistors form another Δ. Any rearrangement you attempt (such as Δ-Y transformation) will 'lose' the currents that you want to find, transforming away the original circuit paths and resistance values.

    Instead, consider using KVL and KCL, or mesh or nodal analysis. There are three loops so three equations will suffice for mesh analysis.
     
  5. Mar 10, 2013 #4
    Oh, I see. In that case, I guess I'll surrender as we haven't covered anything past the regular simplifying method on our lessons and it seems next to impossible to master all those right methods now, in some half an hour - I guess the prof wouldn't be too mad at anyone not having done that, then :)

    Thank you both again very much!
     
  6. Mar 10, 2013 #5

    gneill

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    Staff: Mentor

    Well, you could go right back to KVL and KCL and apply them directly. It'll be a bit of work with a few more equations, but you'll get to the answer.
     
  7. Mar 10, 2013 #6
    OK, I almost did it! I transformed it to a star-like shape and calculated everything by looking at equivalent resistors made in parallel and the only thing left is the current beside the 3 Ohm resistor. Could someone please help? So close yet so far - I can't think of any way to transform it any more so that 3 Ohm is in some more suitable place. Could someone please help how come you get some 0.17A there?
     
  8. Mar 10, 2013 #7

    gneill

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    Staff: Mentor

    Can you show your transformed circuit diagram and describe the steps you took? We need something to work from if we are to help.
     
  9. Mar 10, 2013 #8
    So I rolled the first triangle into something like this:
    http://i.imgur.com/EC0HnMl.jpg

    Thanks to that I had only parallel resistors and series - I simplified the parallel ones and then had the equivalente resistance of the whole circuit thus getting the first and last amperage. The ones beside 2 Ohm and 4 Ohm pair were then easily obtainable as the sum of the current going in and simple proportion. Same goes for the ones beside 6 and 4 Ohms and using the picture above (after tranformation). The one beside 3 Ohm doesn't appear anywhere, though and I can't figure a way to calculate it. Could someone please help?
     
  10. Mar 10, 2013 #9

    gneill

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    Staff: Mentor

    Take the currents that you've calculated and place them on the original circuit. Use KCL to find the missing current.
     
  11. Mar 10, 2013 #10
    Thank you but how can I apply KCL if all the current calculations sum up to the total of 1.125A as of now, without the need to add anything? I got 0.375A at the 4 Ohm resistor and 0.75A at the one above it (slightly off according to the simulation but I guess it's calculated right) - so we have 1.125. Same goes for the ones on the right where I got 0.5 and 0.625 so it sums up, too. What should I sum up, then? 0.375 and 0.5 and add the remaining 0.25? But why these two?
     
  12. Mar 10, 2013 #11

    gneill

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    Staff: Mentor

    If KCL alone doesn't suffice, then apply KVL to find the "missing" potential drops across other resistors, and hence their currents. You said that you have the currents through the 4 and 6 Ohm resistors, and the total through the 5 Ohm resistor. So you know the potential drops on those resistors. Use KVL to find the other potential drops and hence the currents through the other resistors.
     
  13. Mar 11, 2013 #12
    OK, thank you :)
     
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