Node method - 2 ideal voltage sources

Discussion in 'Homework Help' started by xxxyyy, Jan 18, 2012.

  1. xxxyyy

    Thread Starter Member

    Oct 7, 2011
    34
    1
    Hi!
    I was trying to find electrical potential of nodes (picture of circuit below), but I have problem. My referent node is "D". Therefore potential of node "B" is 12V. In this circuit there are 4 nodes, 2 of them are now known ("D" and "B") and two of them I have to find ("A" and "C"). My equation for node "A" is:-I1-I2-I3=0. Current I1 is current from branch with voltage source E2. Problem is,how to represent I1 using Ohm's law ((electrical potential difference)/Resistance) when there is no resistance? If there was resistor connected in series with E2 it wouldn't be problem to write equation, but there is no resistor so what should I do? I hope that someone will help me, it's very important :)
    http://imageshack.us/photo/my-images/546/9670960.jpg/
     
  2. praondevou

    AAC Fanatic!

    Jul 9, 2011
    2,936
    488
    Well, if it's very important...

    Superposition will do the job.

    Short E1, that puts R1 in || with R3 and R2 in || with R4. Both pairs are in series. Calculate current for R3.

    Then short E2, that puts R1 and R2 in || and R3 and R4 as well. Both pairs are then in series. Calculate current for R3.

    Add currents for R3.

    Obtain voltage at A.
     
    anhnha likes this.
  3. xxxyyy

    Thread Starter Member

    Oct 7, 2011
    34
    1
    Is it possible to solve it using node method? :)
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,963
    1,098
    Yes, you can use node method to solve any circuit you have.
    But you have supernode in your circuit.
    So you treat A and C as a one big (super) node.

    Current into the node (entering) is equal current leaving the node.

    I2 + I3 + I5 + I6 = 0 (All the current entering our supernode, and we ignore I1)

    I2 = (Vb - Va)/R1

    I3 = (0 - Va)/R3

    I5 = (Vb - Vc)/R2

    I6 = (0 - Vc)/R4


    And one additional equation

    Va - Vc = E2

    (12V - Va)/125Ω + (0V - 12V)/500Ω + (12V - Vc)/250Ω + (0V - Vc)/500Ω = 0

    and Va = ( Vc + 8V )
     
    anhnha likes this.
  5. Bassalisk

    New Member

    Jul 12, 2011
    8
    1
    You could do it through node analysis, but you have a problem.

    You already found out about that problem, but I will say it again, just to make sure.

    Branch AC has only voltage source, meaning ti has no resistance meaning it has infinite conductivity. You have to compensate for that.

    How can you solve this?

    One way is to treat that into "one big node" as the the poster above said.

    But then again, you can transform this circuit, legally, and get a new circuit which is equivalent to the old one. This includes the necessity to transform a branch BD.

    [​IMG]

    Picture above says it all. I want you to try this method, and can you give us the reference results, so I can do this and check my answers? {I am too lazy to calculate this :( }
     
    anhnha likes this.
  6. xxxyyy

    Thread Starter Member

    Oct 7, 2011
    34
    1
    @Jony130, Bassalisk
    I will try both ways to solve it. Thanks for replies :)
     
Loading...