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Homework Help: Help w/Current Through Batteries

  1. May 6, 2010 #1
    In the figure below, the resistances are R1 = 1.1 , R2 = 1.5 , and the ideal batteries have emfs 1 = 2.0 V, and 2 = 3 = 5.5 V.


    I feel like I've tried everything to solve this problem; I just don't get it!
  2. jcsd
  3. May 6, 2010 #2


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    Use Kirchhoff's circuit laws to solve this. There are more than one way to do this problem, even when using Kirchhoff's laws. But no matter which way you go about it, you will end up with 2 equations and 2 unknowns. Solve the simultaneous equations and things will fall into place after that. :wink:

    [Edit: Alternately, you might be able to use Thévenin/Norton equivalent transformations to grind your way to an answer. But I still suggest using Kirchhoff's laws if you have the choice.]
    Last edited: May 6, 2010
  4. May 6, 2010 #3
    Wouldn't you have three unknowns? The current through R2, the current through R1 in the left loop, and the current through R1 in the right loop? That's what's been confusing me.
  5. May 6, 2010 #4


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    Yeah, if you do it that way, you'll have three unknowns and three equations. I'm guessing you're having problems coming up with that third equation. Am I right? Show us what you've gotten so far.
  6. May 6, 2010 #5


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    Of course you get to define your current loops yourself.

    But let's suppose that you defined your current loops to be i1 on the left side of the circuit, and another current loop i2 on the right side of the circuit. And, let's assume that you have defined both loops to go around clockwise. (These definitions are up to you, btw.)

    Then by that definition, the current going through R2 from the direction of top to bottom, is simply i1 - i2. Or if you are tracing the current from bottom to top, the current is i2 - i1. The important thing is that either way you can express the overall current through R2 as a simple function of the two unknowns. There is no reason to have a third unknown.

    Of course the particular resistor or resistors, who's current is a function of multiple unknowns completely depends on how you define your current loops at the beginning. And the details are a function of how you define the direction of the loops (clockwise or counterclockwise). The current loop definitions I used in the above paragraph is just one possible definition of many.
  7. May 6, 2010 #6
    Okay, so, defining the current to be traveling in a clockwise direction:


    I feel like a third variable is necessary. Because the R1's on either side aren't equal.
    THere is one current going through the left, another through the right, and another through the middle.
    I set a third equation to be i2+i1-i3=0.
    But I wouldn't know where to go from there
  8. May 6, 2010 #7


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    Well, that's not the way I would have done it. But let's use your method. I should still point out though that there is no need to define i2 at all, the way that you've defined it. i2 isn't even a loop. It starts and ends, but doesn't loop back on itself. It is unnecessary.

    But let's go ahead and use it anyway, even if it is unnecessary. But make sure you draw the direction of i2 on your schematic. According to your second and third equation, i2 points up.

    So there is an error in your first equation! If you keep things in terms of i2 which points up, and sum the voltage traced out by following i1 around, the voltage across R2 should get a positive sign, not negative (because you're going in the opposite direction of i2).

    [tex] E_1 - i_1 R_1 + i_2 R_2 -E_2 -i_1 R_1 = 0 [/tex]

    Using your method, solve for i2 (i.e. i2 = i3 - i1). Then anytime you see an i2 in any of your first two equations, substitute the results in, which leaves only i1 and i3! That gives you two equations and two unknowns! :cool:

    But before I finish, I want to point out that using a third variable, such as i2 in this case, can make things more confusing than they need to be, thus making the solution more error prone.

    Based on your loop definitions of i1 and i3, I suggest starting over and ignoring i2 altogether. Trace through the loops in your circuit, and look at the equations that I came up with. Can you see how I did it? (Hint: it didn't involve a third variable or a third equation at all.)

    Left Loop:
    [tex] E_1 - i_1 R_1 - (i_1 - i_3)R_2 -E_2 -i_1 R_1 = 0 [/tex]

    Right loop:
    [tex] -E_3 - i_3 R_1 +E_2 -(i_3 - i_1) R_2 - i_3 R_1 = 0[/tex]
  9. May 6, 2010 #8
    Thanks, that actually helped alot!
    And I looked over your method again, and it does make more sense.
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