KVL Question

Discussion in 'Homework Help' started by hacker804, Mar 3, 2015.

  1. hacker804

    Thread Starter New Member

    Mar 3, 2015
    Hi there.I am new to this forum and recently started a linear circuit analysis course.I have a question regarding Kirchhoff's Voltage Law.Consider the circuit below:
    I need to find the current in this circuit.I can apply KCL and write the loop equation as:
    Now I know that V30=30I and Va=15I

    However the answer in the textbook comes out to be 8A.What did I do wrong?
    Can anyone kindly help?.Thanks
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Va is -15I
    hacker804 likes this.
  3. Marco foschi

    New Member

    Mar 4, 2015
    so based on my calculation I have i= - 8

    Is it correct?
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    No your sign is incorrect.

    Go back to your very first equation. The very last term on the RHS is -Va. If Va is -15*I then -Va=+15*I.
  5. james hudson

    New Member

    Mar 4, 2015
    i am stuck as
    well on this.
  6. hacker804

    Thread Starter New Member

    Mar 3, 2015
    @t n k can you please tell how Va=-15I ?
  7. hacker804

    Thread Starter New Member

    Mar 3, 2015
    Can you please tell how Va=-15I ?
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
    The reason is exactly the same as in the case of -120V. Do you understand why you use -120V
  9. WBahn


    Mar 31, 2012
    Here's a novel idea -- label your diagram and start from basics. You've shown enough of an attempt for me to justify showing you how to do it explicitly because I think that will help you the most.


    KVL says that the sum of the voltage drops around the circuit is equal to zero. The voltage drop from point 'a' to point 'b' is denoted by Vab which equals (Va-Vb). With that in mind, we have

    Vag + Vba + Vcb + Vgc = 0

    We also have the following:
    Vag = 120V
    Vab = I·30Ω = -Vba
    Vbc = 2VA = -Vcb
    VA = Vgc = -Vcg = -I·15Ω

    Plugging these in we get

    Vag - Vab - Vbc + VA = 0
    Vag - Vab - 2VA + VA = 0
    Vag - Vab - VA = 0
    120V - I·30Ω + I·15Ω = 0
    120V - I·15Ω = 0

    I = 120V/15Ω = 8A

    Now let's check our work.
    Vab = (8A)(30Ω) = 240V
    VA = -(8A)(15Ω) = -120V

    If we set Vg = 0V, then
    Va = 120V
    Vb = Va - Vab = 120V - 240V = -120V
    Vc = -VA = 120V

    That means that the voltage of the dependent source is

    Vbc = (-120V) - (120V) = -240V
    Vbc = 2VA = 2(-120V) = -240V

    These agree, so our answer is correct.

    Doing a check is important. I actually made a mistake in one of my equations and originally got an answer of I=2.67A and only caught it because partway through the check it became apparent that I was going to get 0V across the dependent source, which would have required that I=0.

    Notice three key things:

    1) I tracked my units throughout my work.
    2) I asked if the answer made sense.
    3) I checked my work to verify its validity.