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Quick Circuit Practice

  1. Jul 30, 2014 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    What is the emf ##(\epsilon)## of the ideal battery in the following figure?

    2. Relevant equations

    ##i_6 = 1.40 A##
    ##R_1 = R_2 = R_3 = 2.00 \Omega##
    ##R_4 = 16.0 \Omega##
    ##R_5 = 8.00 \Omega##
    ##R_6 = 4.00 \Omega##

    3. The attempt at a solution

    We can find the potential right away:

    ##V_6 = i_6R_6 = (1.40 A)(4.00 \Omega) = 5.6 V##

    Note that ##i_5 = i_6## since ##R_5## and ##R_6## are in series, so we can find the other potential:

    ##V_5 = i_5R_5 = (1.40 A)(8.00 \Omega) = 11.2 V##

    Now we note ##V_4 = V_5 + V_6 = 5.6 V + 11.2 V = 16.8 V## since the resistors are in parallel. Now ##i_4## can be found:

    ##i_4 = \frac{V_4}{R_4} = \frac{16.8 V}{16.0 \Omega} = 1.05 A##

    By KCL once again, we have ##i_2 = i_4 + i_5 = 1.05 A + 1.40A = 2.45 A##. So that:

    ##V_2 = i_2R_2 = (2.45 A)(2.00 \Omega) = 4.90 V##

    Now here is where I have a question. Would it be the case that ##R_2## and ##R_4## are in series? Then I would have ##R_3## in parallel and:

    ##V_3 = V_2 + V_4 = 4.90 V + 16.8 V = 21.7 V##

    So that ##i_3 = \frac{V_3}{R_3} = \frac{21.7 V}{2.00 \Omega} = 10.85 A = 10.9 A##.

    Appyling KCL again yields ##i_1 = i_2 + i_3 = 2.45 A + 10.85 A = 13.3 A##.

    So we get ##V_1 = i_1 R_1 = (13.3 A)(2.00 \Omega) = 26.6 V##.

    Appling the loop rule to the leftmost loop, we obtain:

    ##\epsilon - V_1 - V_3 = 0 \Rightarrow \epsilon = V_1 + V_3 = 26.6 V + 21.7 V = 48.3 V##.

    Hence the emf is ##48.3 V##.
     

    Attached Files:

  2. jcsd
  3. Jul 30, 2014 #2

    SammyS

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    No. They're not in series, but applying the loop rule will give you ##\ V_3 = V_2 + V_4 \ .##


    Other than that it looks good.
     
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