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Circuit problem

  1. Feb 7, 2005 #1
    I have a circuit problem I'm having trouble with. I've attached an image of it.

    [tex] Emf_1 = 3 V; Emf_2 = 1 V; R_1 = 5 \Omega; R_2 = 2 \Omega; R_3 = 4 \Omega[/tex]

    Both batteries are ideal. a.) What is the rate at which energy is dissipated in [tex]R_1[/tex]?

    Since [tex]P = i^2R[/tex], I figured that I should find the current. I used the loop rule (I'm not sure if I should have done this or not). I started at a point between [tex]emf_1[/tex] and [tex]R_3[/tex]. I pretended the current was pointed downward at [tex]emf_1[/tex] and upward through [tex]R_1[/tex] and [tex]R_2[/tex]. I called those last two currents [tex]i_1[/tex] and [tex]i_2[/tex], respectively. [tex]i_{total} = i_1 + i_2[/tex].

    For the inside (left) loop:

    [tex]R_3i_{total}+R_1i_1+emf_1 = 0[/tex]

    For the outside loop:

    [tex]R_3i_{total} + R_2i_2-emf_2+emf_1 = 0[/tex]

    [tex]+R_3(i_1+i_2)+R_1i_1 = -emf_1[/tex]

    [tex]-R_3(i_1+i_2)-R_2i_2+emf_2=-emf_1[/tex]

    [tex](5\Omega)i_1-(2\Omega)i_2 = -7 V[/tex]

    [tex]-(4\Omega)(i_1+i_2)-(2\Omega)i_2 = -3 V[/tex]

    [tex]-(4\Omega)i_1-(6\Omega)i_2 = -3 V[/tex]

    I solved the system of equations and ended up with [tex]i_1 = 0.95 A[/tex]. This isn't the correct answer, because [tex]i^2R[/tex] ends not equalling 0.346 W, which is the right answer.

    I have the feeling I did something really wrong. Did I make it way too complicated? I have a tendancy to overdo things :smile:

    ~Silimay~
     

    Attached Files:

  2. jcsd
  3. Feb 7, 2005 #2

    jamesrc

    User Avatar
    Science Advisor
    Gold Member

    Hi,

    Two things I can see:

    - With the way you defined things and the way you are going around your loops, the signs on your voltages should be switched (e.g. first loop eq should be R3itotal1+R1ii = 0).

    - I think you made some algebra mistakes after that.

    Try again and I think you'll get it.
     
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