Nodal analysis - w/ dependent current/voltage source

Discussion in 'Homework Help' started by castrogfx, Apr 15, 2012.

  1. castrogfx

    Thread Starter New Member

    Apr 13, 2012
    4
    0
    Hi, I again. I'm studying, but I really stop in the basics (and important) things.

    First problem:
    [​IMG]
    Using the nodal analysis, what is the power in the resistor of 4 ohm?
    Answer: 64 W

    ///


    1 (leftest) - i_1 = i_2 + i_3

    2 - 4 V

    3 -  i_4 + i_6 + i_5= i_7

    And the great problem for me (I think):
    What is v_1?
    I put v_1 = -4 V

    i_1 = 2A <br />
 i_2 = \frac {v_1 - 4}{2} <br />
 i_3 = \frac {v_1 - v_3}{4} <br />
 i_4 = 4 - v_3 <br />
i_5 = i_3 = \frac {v_1 - v_3}{4} <br />
i_6 =3v_1 = -12A<br />
i7 = \frac {v_3}{2}<br />
<br />
<br />

    So...


    1 - 2 = \frac {v_1 - 4}{2} + \frac {v_1 - v_3}{4}

    3 - 4 - v_3 - 12 + \frac {v_1 - v_3}{4} = \frac {v_3}{2}

    I have this system of linear equations:
    16 = 3v_1 - v_3<br />
-32 = -v_1 + 7v_3<br />
<br />
v_1 = 4 V   ,   v_2 = -4 V<br />


    what is the power in the resistor of 4 ohm?

    <br />
i_3 = \frac {v_1 - v_3}{4} = \frac {4 + 4}{4} = 2A<br />
P = i i R ->P = 2*2* 4 = 16W



    ////

    And a little help in this:
    [​IMG]

    So, how I can write v_1?


    Final answer (if help): v = -2 V, v_1 = -4 V


    I really appreciate your support!
    Thanks!
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,751
    4,799
    No problem with these, except that this is direct application of KCL. In most places, the term "Nodal Analysis" or "Node Voltage Analysis" refers to a very formalized way of systematically applying KCL such that a set of simultaneous equations can be written down by inspection (or nearly so).


    How did you come up with -4V? The voltage across a current supply is an unknown that you have to solve for. More specifically, it is one of the two node voltages that you are trying to solve for.
     
    castrogfx likes this.
  3. castrogfx

    Thread Starter New Member

    Apr 13, 2012
    4
    0
    Ok. So v_1 = V_1 (the first node)
    And the things go correct right now!
    Thanks for the fast help!

    ///
    I'm still with one doubt (about problem 2).

    [​IMG]

    How I can write v_1?
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,751
    4,799

    Precisely as it is defined; namely, the voltage at the (symbolic) positive node minus the voltage at the (symbolic) negative node. Hence, it will be written in terms of the difference of two of your unknown node voltages. In this case, it will be (v_A - v_B) where 'A' is the node above the resistor and 'B' is the node below it.
     
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