help for my assingment ?

Discussion in 'Homework Help' started by ubeyd, Nov 30, 2014.

  1. ubeyd

    Thread Starter New Member

    Nov 30, 2014
    4
    0
    Hi
    i am a student of computer engineering but my college so insisted to teach me physics :p
    so i have an assignment..i think questions are pretty easy but i am stuck at few points

    so at first question i know how to find the voltage drops if there is only voltage source but there are 2 in this question so any tips ?
    and for the second part, i am familliar with KVL but again 2 voltage source so i am not sure how it effects the answer.

    about the second question i take any advice cuz i am not familliar with Thevenin equivalent yet

    thanks in advance
     
    • 1.jpg
      1.jpg
      File size:
      167.1 KB
      Views:
      56
  2. ubeyd

    Thread Starter New Member

    Nov 30, 2014
    4
    0
    i just need some tips come on guys
     
  3. dalam

    Member

    Aug 9, 2014
    58
    6
  4. MrChips

    Moderator

    Oct 2, 2009
    12,447
    3,363
    Well guess what?

    You cannot become a competent engineer if you do not know physics and math.
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Q1 - simply try applying KVL around the loop you have in the circuit.
    Q2 - use a circuit theory and find Vab voltage, and Vab = Vth, next short A and B together and solve for Isc.
    And finally Rth = Vth/Isc.
    For example we could find Vth by using the superposition for I current.
    I = 10V/(2000 + 200) - 9*I * 200/(2000 +200)
    I = 10/2.2 - 9*I*0.2/2.2
    2.2*I = 10 - 9*I*0.2
    2.2*I = 10 - 1.8*I
    4*I = 10
    I = 10/4
    I = 2.5 and this result is in mA
    I = 2.5mA

    So Vab = Vth = 10V - 2.5mA*2000 = 10V - 5V = 5V
    http://forum.allaboutcircuits.com/threads/thevenins-theorem.102247/#post-771380
     
    • 5.PNG
      5.PNG
      File size:
      56 KB
      Views:
      34
    Last edited: Nov 30, 2014
  6. ubeyd

    Thread Starter New Member

    Nov 30, 2014
    4
    0
    Thank you so much ! you are a life saver
     
  7. dalam

    Member

    Aug 9, 2014
    58
    6
    5.PNG
    In the circuit that you uploaded if I take Vb as a reference node then Va becomes zero.
    I1=0. And I=10/2000. Isc= 10*I=1/20.
    What is wrong in this analysis?
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Nothing is wrong, Isc is equal to 1/20 = 50mA
    So tell me why you think that your analysis is wrong ?
     
  9. dalam

    Member

    Aug 9, 2014
    58
    6
    Untitled.png
    V/2000+V/200=9I+1
    I=-V/2000
    So, V/2000+10V/2000=-90V/2000+1
    101V/2000=1
    V=2000/101
    V/I(test source=1A)=Rth=2000/101
     
  10. dalam

    Member

    Aug 9, 2014
    58
    6
    You got I =2.5mA that's why. Moreover I did not understand how did you use superposition here.
    Can you please testify the answer that I posted in the post above and the approach as well.
    Thanks in advance.
     
  11. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Please check your math again because the correct answer for V is 100V
    And Rth = 100V/1A = 100Ohm
    http://www.wolframalpha.com/input/?i=V/2000 +V/200=-9V/2000 +1

    I got the same result using different approach.
    Vth = 5V and Isc = 50mA so Rth= 5V/50mA = 100 Ohm

    Where I wrote that Isc is 2.5mA ??
    In my post 5 I solve for Vth, but I use superposition to find I current first.

    What you do not understand. I simply use a superposition to find "I" current
    First I remove current controlled current source from the circuit.
    So I' = 10V/(2000 + 200)
    Next I remove 10V voltage source and use a current divide rule.
    I'' = - 9*I * 200/(2000 +200)
    and finally
    I = I' + I'' = 10V/(2000 + 200) - 9*I * 200/(2000 +200)
    Also please read this
    http://forum.allaboutcircuits.com/t...rrent-and-voltage-sources.101488/#post-764893
     
  12. dalam

    Member

    Aug 9, 2014
    58
    6
    Thanks a lot. My bad, I wrote -90V/2000 instead of -9V/2000. And thanks for explaining about superposition. I was not sure weather dependent sources can also be analysed using superposition.
     
  13. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    Yes they can. As long as the circuit is linear, superposition applies (in fact, that is the definition of a linear system).

    Imagine that you have a current I0 that is a function of three different independent sources:

    I0 = I1 + I2 + I3

    Now imagine you have a dependent source that is

    Ix = kI0

    Well, that works out to

    Ix = k(I1 + I2 + I3) = kI1 + kI2 + kI3

    and so we see that it is the sum of the currents that are output by the dependent source in response to the three independent sources individually.
     
  14. dalam

    Member

    Aug 9, 2014
    58
    6
    Thank you, WBahn for helping me out.
     
Loading...