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Parallel circuits, Emfs and kirchoff's laws

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data

    YF-26-62.jpg

    What must the emf ε in the figure be in order for the current through the 7Ω resistor to be 1.79A ? Each emf source has negligible internal resistance.

    2. Relevant equations

    V=IR
    Kirchoffs junc and loop rules

    3. The attempt at a solution

    I put that the outer loop would be 24V +3*I = 7*1.79
    and from there solved for I
    i got that I would be -5.02A
    I then said that because the current I split at the top middle junc the current I{2} down the centre line was I{2}=I-I{1}
    and basically used V=IR to solve for ε which i got to be -13.62V



    but i cant help but feel I've done it wrong, and in the text book there is no answers can anyone help? much appreciated
     
  2. jcsd
  3. Nov 20, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    First, check the equation that you wrote for the outer loop. If the current is assumed to be flowing clockwise then there should be voltage DROPS across both the 3.00 and 7.00 Ohm resistors.

    You might want to consider using nodal analysis (KCL). After all, there's only one node and you already know it's voltage: 1.79 x 7.00. You'll end up with a single equation with one unknown.
     
  4. Nov 20, 2011 #3
    Okay so I got (using kirchoffs lecturer says we have to use kirchoffs laws)
    outer loop :-
    24-3I-7I2=0
    left hand loop :-
    24+ε-3I-2I1=0
    right hand loop:-
    -ε+2I1-7I2=0

    i went on to solve and didnt get a consisted answer... (when put back into equations to check)
    are these equations on the right line? im not the best at circuits.
     
  5. Nov 20, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    I see that you're having trouble keeping the signs straight for the voltage rises and drops whe you "walk" around a loop. Your equation for the outer loop looks okay, but the other two have some problems.

    Take your figure and draw in the assumed directions of each of the currents (I, I1, I2). Now do the left loop again, walking around the loop clockwise starting at the - terminal of the battery. The next component you come to is the source ε. As you walk over it, will there be a voltage rise or a voltage drop?
     
  6. Nov 20, 2011 #5
    Thanks so much i have a consistant answer now :D and i am more confident with circuits.
     
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