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!

Is my work correct?

  1. Jan 29, 2010 #1
    hoe to deal with the dependent voltage source?


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

    For the following circuit vs = 100 cos(1000 t + 20o) V.
    Case 1:
    If the load impedance for the circuit is adjusted until maximum average power is
    delivered to the load:
    a) Find the impedance that should be connected in this case.
    b) Find the maximum average power delivered to the load in this case.
    Case 2:
    If the load is a pure resistive:
    a) Find the value of the load resistor that will maximize the average power in the
    load.
    b) Find the maximum average power delivered to the load in this case.
    Case 3:
    If the load is replaced with a series combination of 5 Ohms resistor and a 5 μF capacitor
    and the one Ohm resistor is replaced with a box called Load2.
    a) Find the impedance that should be connected to the location of load2 to
    maximize the average power in Load2.
    b) Find the maximum average power delivered to the load in this case.
    c) If Load2 is a pure resistive element, what should be the value of this load to
    maximize the average power in Load2 and what is the value of the maximum
    average power in this situation.


    2. Relevant equations



    3. The attempt at a solution
    Case I:
    *The value of the 10 mH conductor = ωjl = 1000*10m = j10Ω
    *The value of the 3 mH conductor = ωjl = 1000*3m = j3Ω

    A)we have to find Thivenin impedance :
    - Applying KVL in the first loop to find (ix):
    - 100∟20+(25+j10) ix + 5 ix =0
    (30+j10) ix =- 100∟20 → i_x=3.162+j0.1=3.162∟1.8 A
    → The dependent voltage source = 15.8∟1.8 V
    - Using source transformation:
    Current source 1 = 3.162∟1.6 A & parallel to (25+j10) Ω impedance
    & Current source 2 = 15.8∟1.8 A & parallel to 1 Ω resistor
    - combining parallel impedances to find equivalent impedance Z_th:
    (25+j10) ║ 1 ║ j3
    → 1/Z=1/((25+j10))+1/10+1/( j3) →Z_th=0.6∟13= 0.585+j0.135 Ω
    → Z_l= Z_th^* = 0.6∟-13= 0.585-j0.135 Ω
    Which will gives maximum power.

    B) finding the complete Thevenin circuit to find the value of the power:
    - combining parallel current sources:
    3.162∟1.6 + 15.8∟1.8 = 19∟4.39 = 19+j1.46 A
    - source transformation again to have Thevnin equivalent circuit:
    → V_th= 11.4∟17.4 V
    → P_max=〖(I/√2)〗^2*R_th = 〖(19/√2)〗^2*0.585 = 105.6 W

    Case II:
    A) The value of the load resistor = √((R_th^2 )+(X_th^2 ))= √(〖(0.585)〗^2+〖(0.135)〗^2 )= 0.6 Ω

    B) The maximum power = (V_m^2)⁄(8R_th )= 〖(11.4)〗^2⁄((8)(0.6))=27 W

    Case III :
    *The value of the 5µF capacitor = (-j )/ωc= - j200Ω

    A) we have to find the Thevinin equivalent circuit:
    - using parallel/series combinations:
    (5-j200) ║ j3 = (j3(5-j200))/(j3+(5-j200)) = 3.04∟89.9 = 0.01+j3.04 Ω
    - using source transformation:
    Current source 1 = 3.162∟1.6 A & parallel to (25+j10) Ω loud.
    & Current source 2 = 5.197∟-88.2A & parallel to (0.01+j3.04) Ω loud.

    - combining parallel impedances to find equivalent impedance Z_th
    (25+j10) ║ (0.01+j3.04)
    →Z_th=((25+j10)(0.01+j3.04) )/((25+j10)+(0.01+j3.04)) →Z_th= 2.9∟84.1 Ω
    → Z_l=2.9∟-84 which will give the maximum power.
    B)Finding the complete Thivenin circuit to find the value of the power:
    - combining parallel current sources:
    3.162∟1.6 + 5.197∟-88.2 = 6.1∟-56.9 A
    - Source transformation again to have Thevnin equivalent circuit:
    → V_th= 17.69∟27.2 V
    → P_max= (V_m^2)⁄(8R_th )= 〖(17.69)〗^2⁄((8)(0.298))= 131.26W
    C) The value of the load resistor = √((R_th^2 )+(X_th^2 ))= √(〖(0.298)〗^2+〖(2.885)〗^2 )= 2.9 Ω
     
  2. jcsd
  3. Jan 30, 2010 #2
    you should attach the picture of the circuit ( I attached for you )

    I obtained different answers

    you should use mesh-current method not the source transformation
     

    Attached Files:

  4. Jan 30, 2010 #3
    Zayer...
    Are you a KFUPM student ?

    but why i can't use source transformation?
     
  5. Jan 30, 2010 #4
    I am your classmate

    Because you will end up with independent source , dependent source and impedance which it not Thévenin equivalent circuit
    Also,you have to keep the impedance with the current Ix
     
  6. Jan 30, 2010 #5
    OMG!!!
    that means i brought the eid!!!

    but
    can't we use it as independent source after finding Ix ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook