1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Circuits-dependent source

  1. Jan 27, 2012 #1
    1. The problem statement, all variables and given/known data
    i attached the problem.
    I am unsure how to solve this problem. I applied KVL and got 15-1I-2vx-vx-2I=0. Where I is the current in the circuit. From that equation I have 2 unknowns, so I obviously have to find I.

    To find I I'd have to find Req, but I don't know how the resistors are arranged. the 2 ohm and 5ohm resistors are in series giving an equivalent R=7ohms. Is the 1ohm resistor in parallel with the combination R=7ohm?
     

    Attached Files:

  2. jcsd
  3. Jan 27, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    It's a series circuit -- all the components are connected in series.
     
  4. Jan 28, 2012 #3
    I don't get how it's a series circuit. Isn't the voltage dependent voltage source (2vx) separating the 1 ohm resistor from being in series with the others?
     
  5. Jan 28, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    No, the dependent source is just another component in the loop. Note how there is only one path for current to follow around the single loop. This is a series circuit.

    attachment.php?attachmentid=43217&stc=1&d=1327761220.gif
     

    Attached Files:

  6. Jan 29, 2012 #5
    Oh I see thanks. Does that dependent source have an effect on calculating the current? For this problem I found Req=8ohms. Would current just be 15V/8ohms=I or do I have to take into account the dependent source?
     
  7. Jan 30, 2012 #6

    gneill

    User Avatar

    Staff: Mentor

    The dependent source certainly has an effect on the current.

    Start by writing the KVL equation for the loop.
     
  8. Jan 30, 2012 #7
    I did and got 15-1I-2vx-vx-2I=0.
    Would my other equation be V=IR, I=(15-2vx)/8
     
  9. Jan 30, 2012 #8

    gneill

    User Avatar

    Staff: Mentor

    I suppose you could use that equation, but why not something simpler, like

    Vx = I*5Ω ?
     
  10. Jan 30, 2012 #9
    Yeah I just realized the Circuit said what the drop was over the 5 ohm resistor
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook