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Circuit with Resistors and Two Opposing EMFs

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data
    The circuit shown in Fig. 25.37 contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit (magnitude and direction) . . .

    The circuit is a rectangle with the emfs at the top and bottom sides and the resistors on the left and right sides. The positive side of each emf is towards the left side, so that one creates a current clockwise and the other an opposing one counterclockwise. The resistances are: left, 5.0; right, 9.0; top (internal), 1.6; bottom (internal), 1.4. The emf at the top is 16 and at the bottom 8.

    2. Relevant equations
    V = IR
    V = E - Ir
    I = E / (r+R)

    3. The attempt at a solution
    For the formula I = E / (r+R), since the denominator is the sum of all the resistances, I assumed I could sum all the emfs as well (taking sign and direction into account). So I did:

    I = (16-8) / (1.4+5+1.6+9) = 8/17

    I do not know what the correct answer is, but I intuitively feel like this answer is wrong.
     
  2. jcsd
  3. Feb 20, 2009 #2
    The current magnitude is correct. How is the current drawn on the circuit, is it flowing clockwise or counterclockwise?
     
  4. Feb 21, 2009 #3
    Wow, so that is the correct magnitude? That's good. The direction should be counterclockwise, since the top emf is stronger than the bottom one, and the top one is going counterclockwise.

    And just for clarity I drew a rough sketch of the diagram:

    PhysicsCircuit.png
     
  5. Feb 21, 2009 #4
    Yes, your magnitude and direction are correct. Just remember that when doing a KVL around a loop, if your answer is positive your assumed current direction is correct. If negative, the direction of the current is opposite of the assumed direction.


    For example, say I was summing the voltages around the loop in the clockwise direction:

    16V + I(9) + (-8V) + I(1.4) + I(5) + I(1.6) = 0V
    I = -8/17 A

    Since I assumed a clockwise direction for the current, the answer came out negative. This is telling me the current is actually going in the opposite of the assumed direction.
     
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