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Direct Current Circuit

  1. Nov 29, 2008 #1
    I have a problem solving the following direct current circuit:

    [​IMG]

    The Resistor-values are know, as is the ems. The question is to define the current in each part of the circuit.

    I thougth of solving it with kirchhoff rules, and therefore I defined several currents. (I, A-> E)
    I set up some equations:
    I= A + B
    B= C + E
    A + C = D
    E + D = I
    Then I thought of using the rule about the potential that is 0 in a loop, but I haven't got a clue how to enter the ems in these 2 equations.
     
  2. jcsd
  3. Nov 30, 2008 #2
    Kirchoff's second rule is applied as follows:

    1. Define a closed loop within the circuit, in one particular direction.
    2. When travelling past an emf source from negative to positive terminal, + emf, if in the reverse direction, - emf.
    3. When travelling past a resistor in the direction of a current, -IR, and against current, +IR
    4. The sum of the terms you have "collected" in a loop is zero, by the Kirchoff Rule.
     
  4. Nov 30, 2008 #3
    But I havent got a clue how to fit the emf into these equations.

    Suppose I define a loop in the upper square, i've got: (clockwise rotation)

    - A . R1 + C . R2 + B . R3 + ? = 0

    And in the lower square i've got: (clockwise rotation)

    - E . R4 - C . R2 - D . R5 + ? = 0

    And if you take the big loop, i would take this equation (also clockwise):

    + emf - I . R (with R = total resistance over the 2 squares)


    Is this correct?
     
  5. Nov 30, 2008 #4
    You got it about right...

    1st eqn is about right, if your loop has no source, emf need not be included, so remove the question mark

    2nd eqn has an error, going clockwise, at E you're going against the current, so it's E . R4.

    3rd eqn is correct, but it helps if you break it down into one particular path through the squares, Like emf - B.R3 - E.R4.

    Also, it helps if you choose loops that minimises the number of different variables you bring in.

    Lastly, it's OK if you are unsure of the direction of current C, as long as you keep its direction consistent in all equations. If you chose the wrong direction, your answer will emerge negative, and you just have to correct the direction.
     
  6. Nov 30, 2008 #5
    Thanks for helping me.

    If you define paths like:
    emf - B.R3 - E.R4 = 0
    emf - A.R1 - D.R5 = 0

    Can you say the following:
    for the first loop: R3 and R4 are in series: so I count them together, and use V = I.R to calculate the current in the loop?
    for the second loop: R1 and R5 are in series, so I count them together and use V = I.R to calculate the current in that loop?

    And with these two currents you can find the current over R2?
     
  7. Nov 30, 2008 #6
    Unfortunately, it isn't as easy as that. B and E are most certainly different, and so is A and D, because C is non-zero. Care must be taken here because the current in a single loop is of most often not constant. However, here's where the first rule comes into play. The number of variables can be reduced by relating the different currents using the junction rule.

    Remember, addition of resistance in series can only be applied in cases of constant current. I believe the source of your confusion is in eqn 3(earlier), that equation is true because in current going into the squares is equal to the current leaving the squares. When breaking down into the constituent paths, the current I, breaks up into A, B and so on, and hence it will not be as easy as adding the resistance in series together. Hope this clears the matter up.

    PS: It would help if you include a question with target variables, and given variables, to help illustrate the points above.
     
  8. Nov 30, 2008 #7
    You cleared it up.

    Well: these are the values:
    V= 2 v
    Resistance at point B = 22 ohm
    " " " A = 39 ohm
    " " " C = 10 ohm
    " " " E = 100 ohm
    " " " D = 82 ohm.

    I put the equations we found in a lineair programming tool to find the following values:
    A= 0,032 Amp
    B= 0,052 Amp
    C = 0,013 Amp
    D = 0,045 Amp
    E = 0,032 Amp
    I = 0,084 Amp.

    But I wouldn't have found them without the tool...
     
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