Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    http://img249.imageshack.us/img249/5889/chapter4problem15hd1.jpg [Broken]

    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].

    http://img180.imageshack.us/img180/5125/chapter4problem15part2nx3.jpg [Broken]

    Expressing the currents in terms of node voltages:


    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.

    http://img263.imageshack.us/img263/5007/chapter4problem15part3jk5.jpg [Broken]

    Express the currents:


    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]


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


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


    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:


    Now, the last source.

    http://img412.imageshack.us/img412/3179/chapter4problem15part4ze2.jpg [Broken]

    Express the currents:


    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:


    EDIT: Thanks for the help chanvincent!!!
    Last edited by a moderator: May 2, 2017
  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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook