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Basic CIRCUIT ANALYSIS question

  1. Feb 1, 2007 #1
    1. The problem statement, all variables and given/known data

    Please view the attached image.

    2. Relevant equations

    P = VI = I^2 R = V^2 / R
    V = IR
    KVL

    3. The attempt at a solution

    I'm having trouble with part A. I got part B.

    My Part A Problem: I chose my reference direction of I to be going clockwise (to satisfy passive sign convention). Now, since P is positive (which denotes absorption), we know that I = .25mA. How? P = IV... 2.5mW = 10V * I.

    Knowing that, I found Vx by using Ohm's law.

    V = IR
    Vx = -.25mA * 10v = -2.5V

    Knowing that, to compute Vba, what I did was start from point B, go towards pt A.

    So...

    Vx + RI + 2Vx = Vba
    (-2.5V) + (3kOhm)(-.25mA) + 2(-2.5) =

    Vba = -8.25 V

    How I solved part B: I think I got lucky here, cause what I did previously is this:

    P = VI = 2Vx * I = 2 (2.5) * .25mA... Notice it's + * + = +.. whereas it could be - * - = +, so my online HW told me it was right. However, now i'm skeptical about the sign of my amp when computing Vba.

    However, when I recalculated my current (I forgot to negate it while goign CCW), I come out with a negative current. So, I'm curious, is my Vba answer correct?
     

    Attached Files:

    Last edited: Feb 1, 2007
  2. jcsd
  3. Feb 1, 2007 #2
    When I did this problem, I had my current's reference direction going counter clockwise (this would be satisfying the passive sign convention being that the voltage drop across the 10k goes from a high potential on the bottom terminal to a lower potential on the top, according to the reference voltage Vx)

    When I solved for the current, I got a negative current, So from then on I just worked the circuit with current going CW. I got the same answer as you. I think you'll be fine as long as you stick with your reference current direction until you get an answer for it, then you'll know whether your chosen direction was correct or not.
     
  4. Feb 1, 2007 #3
    I see what you did, thanks for working the problem out. So, do you think that my Vba is correct?
     
  5. Feb 1, 2007 #4
    The first part is correct. For part b, the power supplied/absorbed by the dependent source is, according to the passive sign convention, 2Vx multiply the current from the +terminal to -terminal. This means 2(-2.5)*(-0.25mA) which gives you a positive answer and that indicates power dissipated.
     
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