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Homework Help: Current through resistors in circuit with TWO ideal emfs

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data
    In the figure, the ideal batteries have emfs 1 = 17.0 V and 2 = 0.5001 V, and the resistances are each 1.22 Ω.

    I tried to include a link but I am unable.

    As best as I can describe: There is one main loop with a wire connecting the bottom and the op piece. The two emfs are on the left and right side BOTH pointing towards the top (positive is both towards the top of the circuit/each other). R_1 is on the left side top and R_2 is on the right side top. R_3 is in the middle piece of connecting wire.

    I hope that explains it well enough.

    a) What is the current in resistor 2?
    b) What is the current in resistor 3?

    2. Relevant equations
    ƩΔV=0 for a complete circuit
    Ʃi in a junction must be equal. This is to conserve charge.

    3. The attempt at a solution
    I am getting close but the double emfs are confusing me. I have another problem similar to this that I am also not able to do. I have the test Monday and I am certain there will be a double emf circuit problem because our professor is just that sort of guy. He also never covered this in class.

    i_2 goes out of the positive terminal of ε_2 into R_2, i_1 goes out of the positive terminal of ε_1 into R_1, and i_3 goes straight down.

    Using the junction rule I get:

    i_3=i_1 + i_2

    I tried using the loop rule for the right loop going in a ccw direction assuming the current is going in the same direction -ccw.

    I get -i_1*(R_2)-i_3*(R_3)+ε_2=0

    left loop: -i_1*(R_1)-i_3*(R_3)+ε_1=0

    I then subbed the three equations into each other using wolfram alpha to check my work before I did it by hand and they are wrong. I am not sure what I am doing incorrect and I tried to follow an example from the book but I could not figure out how to apply that material to this case.

    Thanks for the help. I will post more later as I attempt it for the third time today.
  2. jcsd
  3. Jun 8, 2012 #2


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    Did you mean i1 in this equation, or was it supposed to be i2?

  4. Jun 8, 2012 #3
    That is supposed to be i2, sorry. The rest of the equations look correct.

  5. Jun 8, 2012 #4


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    Your equations sound okay to me then, if I'm picturing the circuit correctly. When I plug the numbers in, I find i2=-4.37 A and i3=4.78 A. What did you get?
  6. Jun 9, 2012 #5
    I wrote a long explanation of my work and then i went to submit and it lost it.

    Basically I worked out the three loops again and simplified using the fact that R is equivilant to get the following system:


    Wolfram alpha says that is impossible to solve. I would really appreciate any advice.

    I have a picture of the circuit but I can not post it. That is making this really difficult to explain.
  7. Jun 9, 2012 #6
    I'm really sorry I did not notice this before. emf1=17V emf2=.5*emf1 or 8.5V. All the resistance is 1.22ohms

    My apologies that I did not catch that, it did not copy correct. I have been using 8.5V for the problem.

    Again, sorry about my mistake.
  8. Jun 9, 2012 #7


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    You seem to have made a sign error somewhere, and you're missing an equation. If you subtract the first equation from the third, you end up with ##i_1 + i_3 = -13.9##, which is clearly inconsistent with equation 2.

    The three equations you should have started with are
    1.22 i_1 + 1.22 i_3 &= 17 \\
    1.22 i_2 + 1.22 i_3 &= 8.5 \\
    i_1 + i_2 &= i_3
    \end{align*} which are just the equations you wrote above with the numbers plugged in.
  9. Jun 9, 2012 #8
    I guess I was making everything to complicated. It works. I'm not sure why I was so confused.

    There is no current through R_2. Why is that? If there is a connection between the emfs should there not be current everywhere in the circuit?

  10. Jun 9, 2012 #9


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    Usually there would be a current through R2, but it just happens to work out that the potential drop in this circuit across R2 is 0, so no current will flow through it.
  11. Jun 9, 2012 #10
    Interesting. Thanks for the help.
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