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

Homework Help: Power in voltage dependent source

  1. Jul 9, 2006 #1
    I have some problems determining the power dissipated in the dependent source,mostly because my equations aren't right, I think...So,I was wondering,if someone could help me getting my equations right...Here's what I've come with so far:


    In the left node point, the voltage is 132v(I think...does one have to add the contribution of the dependet source?)
    Setting the bottom node as the node reference I'm only left with 2 nodes,so,the equations for those would be:

    Eq. 1: (v1-132)/7 + (v1-v2)/3 + v1/5=0
    Eq. 2: v2/10 + (v2-v1)/3 + (v2-132)/2=0

    http://i75.photobucket.com/albums/i281/esmeco/dependentsource.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jul 9, 2006 #2
    I don't see how you can get 7I0 = 132. And I think Eq. 2 is wrongly formulated. Also, it would help out if the labels (node 1, 2, etc.) that you used are explicitly illustrated to avoid any confusion.
  4. Jul 10, 2006 #3
    The current flowing through the 2-ohm resistor is not (v2-132)/2.
  5. Jul 10, 2006 #4
    Here are the labelled nodes...I'm not sure if I should put an extra node on the right node that isn't labelled,but I guess not since the voltage on that node is already known(1329...

    http://i75.photobucket.com/albums/i281/esmeco/dependentsource2.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  6. Jul 10, 2006 #5
    The voltage on that node is not known. It's not 132V. It's 7Io; the voltage depends on the current Io, which is through the 10-ohm resistor. So you have to obtain an equation for Io in terms of V2.
  7. Jul 10, 2006 #6
    So,the 132v doesn't influence on that node voltage?Since the 132 voltage source is connected to that node,shouldn't we also add the 132v to the 7I0 ?Wouldn't the voltage on that node be something like 132 + 7I0?
    So,the equation for I0 in terms of V2 would be I0=V2/10...
  8. Jul 10, 2006 #7
    132V + 7Io is the voltage of the node to the left of the 7-ohm resistor (since you didn't label it, let's call this node B). The voltage of the node to the left of the 2-ohm resistor is 7Io (let's call this node A). The 132V voltage source increases the electric potential (that is, "voltage") from A to B, so it actually means that the DIFFERENCE in electric potential between the two nodes is 132V.
  9. Jul 10, 2006 #8
    So if i get it,the equations would be:

    Node B -> 132 + 7Io
    Node A -> 7Io
    Node v1 -> (v1-132)/7 + (v1-v2)/3 + v1/5
    Node v2 -> V2/2 + 7Io + v2/10 + (v2-v1)/3

    I'm not quite sure if those v1 and v2 equations are right...The dependet source and voltage source both connected to node A still confuse me a bit...But thanks anyway for the replies!
  10. Jul 10, 2006 #9
    No you're not getting it... Equation 2 is still wrong. You're confusing current and voltage. 7Io is NOT the current through the dependent source. Look at its circuit symbol. It's a CURRENT-DEPENDENT VOLTAGE SOURCE. That means it's actually a VOLTAGE SOURCE, which depends on CURRENT.

    Okay, you're confused about node A. Let's forget about voltages for a minute and consider a gravitational analogy. You know from physics what gravitational potential energy is, right? The higher a certain object is, the higher its potential energy. That means, when released, it will come down with a kinetic energy which depends on the difference between the initial potential energy and the final potential energy, correct?

    Now let's come back to electrostatics. Electric potential can be thought of as gravitational potential, and charge can be thought of as the object I mentioned. "Current" is the movement of charges between two points.

    Now look at your circuit. We can think of the voltage sources as elevators. From your ground node (incidentally, the ground node always has 0 V) to node A, there is the 7Io voltage source, correct? You can think of the 7Io source as bringing you up to, say, level 3 in a building. And then from node A to node B, there is the 132 V source brings you from level 3 to a higher level, let's say level 5. Get it now?
    Last edited: Jul 10, 2006
  11. Jul 11, 2006 #10
    So,since 7Io is the voltage of the dependent source we sould add it on equation 2 like this?:

    Eq. 2:(v2-7Io)/2 + v2/10 + (v2-v1)/3=0
  12. Jul 11, 2006 #11
    That's correct.
  13. Jul 11, 2006 #12
    Thanks!Still, when solving the equations,the solution I get for the power doesn't match with the one provided!I have the voltage on the dependent source after determining Io and what i need to know now is the current that enter and leaves the A node so I can get the current on the dependent source,right?
    These are the steps I took to find the values of v1 and v2:

    {(v2-7v2/10)/2 + (v2-v1)/3 + v2/10=0 <=> {(10v2-7v2)/20 + (v2-v1)/3 + v2/19=0

    <=> { 9v2/60 + (20v2-20v1)/60 + 6v2/10=0 <=> { 35v2=20v1 <=>

    { v1=35v2/20

    { (15v1-1980)/105 + (35v1-35v2)/105 + 21v1/105=0 <=> { 71v1 - 35v2=1980 <=>

    { 71(35v2/20) - 35v2=1980 <=> { (2485v2 - 700v2)/20=1980 <=> { v2=22.1V

    {v1= (35x22.2)/20 = 38,8V

    Io=22.1/10 =2,21A
    7Io= 2.21x7 =15.47V
    Vdependetsource= 15,47V
    Idependentsource= (38,8-132)/7 + (22.1 - 15,47)/2 = 10A

    So,the power of the dependent source would be:

    Pdp= 15,47x10=154,7W

    I'm not sure if these steps are right because the power doesn't match with the one provided in the solution,but the solution could also be wrong...Anyway if someone could tell me if these steps and values are right it would be immensely appreciated!
    Last edited: Jul 11, 2006
  14. Jul 12, 2006 #13

    Anyone?I'm getting a bit desperate...:frown:
  15. Jul 13, 2006 #14
    I am not sure if the answer is given correctly. I got V1 = 44, V2 = 176/7, Io = 88/35 and P = -7Io * (Io+V1/5) = -199.1W when I tried to work it out.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook