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CIRCUIT ANALYSIS: Use superposition to find the current I

  1. Jan 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Use superposition to find I in the circuit below.

    [​IMG]


    2. Relevant equations

    KCL, KVL, Super-position principle, V = i R


    3. The attempt at a solution

    I turn only one source on at a time and solve for [itex]I_1[/itex] through [itex]I_3[/itex].

    [​IMG]

    Expressing the currents in terms of node voltages:

    [tex]I_1\,=\,\frac{V_2}{3}\,\,&\,\,I_2\,=\,\frac{V_1}{2}\,\,&\,\,I_3\,=\,\frac{V_1\,-\,V_2}{1}\,\,&\,\,I_4\,=\,\frac{V_1}{4}[/tex]

    KCL @ V1) [tex]I_2\,+\,I_3\,+\,I_4\,=\,-2\,\longrightarrow\,7\,V_1\,-\,4\,V_2\,=\,-8[/tex]

    KCL @ V2) [tex]I_3\,+\,2\,=\,I_1\,\longrightarrow\,3\,V_1\,-\,4\,V_2\,=\,-6[/tex]

    Using those two equations in two variables, I get that [itex]V_2\,=\,\frac{9}{8}\,V[/itex].

    Substitute back into equation for[itex]I_1[/itex] and I get [itex]I_1\,=\,\frac{3}{8}\,A[/itex]


    Now, to the second source.

    [​IMG]

    Express the currents:

    [tex]I_1\,=\,\frac{V_1}{2}\,=\,-10\,A\,\,&\,\,I_2\,=\,\frac{V_3}{3}\,\,&\,\,I_3\,=\,\frac{V_2\,-\,V_3}{1}\,\,&\,\,I_4\,=\,\frac{V_2}{4}[/tex]

    KCL @ V2) [tex]I_1\,+\,I_3\,+\,I_4\,=\,0\,\,\longrightarrow\,\,6\,V_1\,+\,3\,V_2\,+\,4\,V_3\,=\,0[/tex]

    KCL @ V3) [tex]I_3\,=\,I_2\,\,\longrightarrow\,\,3\,V_2\,-\,4\,V_3\,=\,0[/tex]

    [tex]3\,V_2\,=\,4\,V_3[/tex]

    Plug this into the equation for KCL @ [itex]V_2[/itex]:

    [tex]6\,V_1\,+\,6\,V_2\,=\,0[/tex]

    We know the voltage difference between [itex]V_1[/itex] and [itex]V_2[/itex]:

    [tex]V_2\,-\,V_1\,=\,20\,V[/tex]

    Solving: [tex]V_1\,=\,-10\,V\,\,&\,\,V_2\,=\,10\,V\,\,&\,\,V_3\,=\,\frac{15}{2}\,V[/tex]

    Now plug the voltages into the [itex]I_2[/itex] v = i R equation:

    [tex]I_2\,=\,\frac{V}{R}\,=\,\frac{\frac{15}{2}\,V}{3\Omega}\,=\,\frac{5}{2}\,A\,=\,2.5\,A[/tex]


    Now, the last source.

    [​IMG]

    Express the currents:

    [tex]I_1\,=\,\frac{V_1}{2}\,\,&\,\,I_2\,=\,\frac{V_1\,-\,V_2}{1}\,\,&\,\,I_3\,=\,\frac{V_2}{3}\,\,&\,\,\frac{V_1\,+\,16}{4}[/tex]

    KCL @V1) [tex]I_1\,+\,I_2\,+\,I_4\,=\,0\,\longrightarrow\,7\,V_1\,-\,4\,V_2\,=\,-16[/tex]

    KCL @ V2) [tex]I_2\,=\,I_3\,\,\longrightarrow\,3\,V_1\,-\,4\,V_2\,=\,0[/tex]

    I get [itex]V_2\,=\,-3\,V[/itex] and I plug into the original current equation and I get that [itex]I_3\,=\,-1\,A[/itex].

    Finally, add them up:

    [tex]I\,=\,I_1\,+\,I_2\,+\,I_3\,=\,\frac{15}{8}\,=\,1.875\,A[/tex]

    EDIT: Thanks for the help chanvincent!!!
     
    Last edited: Jan 26, 2007
  2. jcsd
  3. Jan 26, 2007 #2
    For your second figure, I spoted a little mistake

    The voltage accross the 2[tex] \Omega [/tex] resister is [tex] V_1[/tex], not -20.

    -20 is the voltage of [tex] V_2 - V_1[/tex]

    If you correct this little error, you will get [tex] I_2 = 2.5 [/tex]V... which will yeild a correct answer.
     
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