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Help New to the circuit

  1. Mar 26, 2008 #1
    Help!!!!New to the circuit

    1. The problem statement, all variables and given/known data
    Question 1
    Use the Mesh analysis to find v
    0 and power absorbed by the 20 resistor in the circuit
    Question 2
    a) Find the Thevenin equivalent circuit external to the resistor R
    b) Find the condition under which maximum power is transferred to R
    c) Determine the maximum power transfer to R

    Does anyone know how to solve them I tried them but it didnt seen right and the dependent source is so annoying is there any way to get rid of them? thanks

    p.s untitiled is question 1 and 2 for question 2 thankz
     

    Attached Files:

  2. jcsd
  3. Mar 26, 2008 #2
    i got Vo + 24v dont know if thats right and quesrion 2 im bit confused i got Rth 10ohms sound right?
    Need great help thankz
     
  4. Mar 29, 2008 #3

    CEL

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    There is no way to cancel dependent sources. Show what you have done in question 1 in order to be helped.
    For question 2. Va is the difference between the 10V source and the open circuit voltage Vab. Develop your analysis and you will have one equation with one unknown: Vab.
    In order to calculate the equivalent resistance, the best approach with dependent sources is to use Rth = Voc/Isc. You already have Voc. Short circuit terminals a and b and calculate Isc.
     
  5. Mar 29, 2008 #4
    i get number one now i have I1 as -2.585A ,I2=-3.732A ,I3=-5A
    AND V0= (I1-I2)R= 23V AND POWER = 26.45W
    But i still dont know question2 where is the 10v u talking about here is and is Vab = Vth?
     
  6. Mar 29, 2008 #5

    CEL

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    Sorry! I meant 12V, not 10V. And Vab = Vth.
     
  7. Mar 29, 2008 #6
    so is Rth = 4+2+4=10?
    and Vth =
    (Vth-2Va)/4 +Vth -12/4 -4=0
    5Vth-4Va=84

    what's Va?
    12V/4=3A
    3+4=7A
    7*4=28v=Va????
    Sorry im trying my best to understand it
     
  8. Mar 30, 2008 #7

    CEL

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    Write the equation for the outside mesh, using Va as unknown. Remember that the current through the resistors 2 ohm and 4 ohm after the controlled source is the sum of the current through the firsr 4 ohm resistor with 4A of the current source. From this I found Va = -24V and Vab = Vth = 36V.
    Next, short circuit the points a and b and calculate the current through them, Isc.

    Rth = Vth/Isc.
     
  9. Mar 30, 2008 #8
    CEL can you write out the equation coz i cant get anywhere close to 36V for Vth!!!!!
    I did node a (Vth-12)/4 -4+ 2Va+Vth/6=0
    Vth equals 14 sth and not even -24 or 36 I am really really lost and i spend any hours on this still cant finish part A how do u do part B and C
    Max power is = Rth as well as Rn but where do i go from there

    damn i feel stupid with this circuits stuff!!!!
     
  10. Mar 30, 2008 #9

    CEL

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    Calling Ia the current trough the 4ohm resistor:
    Ia = Va/4
    The outermost mesh:
    -12 + Va - 2Va +(2+4)(Ia + 4) = 0
    or
    -12 -Va + 6Va/4 + 24 = 0
    12 + Va/2 = 0
    Va = -24V.

    Vth = 12 - Va = 12 + 24 = 36V.
     
  11. Mar 30, 2008 #10
    Wow are u a teacher? That was quick so u did a supper mesh and find out Va and Vth is an open circuit ya? and Now how do i find R's resistor Rth is a short circuits right? only way i know is Rth=Vth/Isc any other way to solve other then the one i menotion any hint to start with finding Isc?

    sorry my head is just not working atm and been on this question like 9 or 10 hrs already and havnt sleep for longlong time


    I alomst finishing my self study!!! YOU should teach me I'm kind of doing it as self study so hard without a teacher
     
  12. Mar 30, 2008 #11

    CEL

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    When you short circuit points a and b you get two independent meshes: one with the two independent sources an one resistor and the other with the dependent source ant two resistors. Work from there.
     
  13. Mar 30, 2008 #12
    k i have loop 1 as -12 +4I1+16=0 and I1 =-1A

    loop2 as -48+6I2=0 then I2=8A

    is this what u mean in the last post and if so Isc= (I1-I2) = -9A
    is Rth become -4?

    max power would be???
     
  14. Mar 30, 2008 #13

    CEL

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    Where did you get the value 16 in the first equation? Loop 1 comprises the 12V source, the 4 ohm resistor and the short circuit ab.
     
  15. Mar 30, 2008 #14
    so u mean loop1 as -12+4I1=0 SINCE SHORT CIRCUIT AB IS THERE?
    I1=3 BUT IS LOOP 2 RIGHT?

    THEN ISC = -5A, Rth = -36/5

    somehow i hva a feeling of thats wrong
     
  16. Mar 30, 2008 #15

    CEL

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    It is wrong. Don't forget that the current source also flows through the short circuit and that Va has now a new value that must be substituted in the dependent source.
     
  17. Mar 30, 2008 #16
    so what do u do with the current source then


    ya VA was sub and loop2 = 2*-24+2i2+4i2=0
    -48+6i2=0
    i2=8A isnt that it?

    and is ISC=(I1-I2)???????? IM KIND OF CONFUSED
     
  18. Mar 30, 2008 #17

    CEL

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    Where did you get the -24 value? Va = 4i1!!!
    You have 3 currents through the short circuit: i1, i2 and 4A from the current source.
     
  19. Mar 30, 2008 #18
    im really confused
    so how many loop is it

    is it one on the lefthand side and second on the right of short circuit
     
  20. Mar 30, 2008 #19
    -12+4i1+16=0 loop 1
    what do i do with the 4A in this loop do u do source converction? and change it to voltage? then thats 16V 4A*4OHMS


    coz VA =-24V AND 2VA=-48V for the dependent source
    2 and 4 ohms are added up
    -48+6i2=0 and this is loop2


    and one thing is Isc=(i1- i2)?????? if so how that make sense i have a negative number for current
    that is how i arrive at that point and im sorry i cant get it any further
     
  21. Mar 30, 2008 #20
    are u saying 4i1+6i2=0 for a second equation?
     
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