1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

Thevenin Equivalent Circuit

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data

    I wish to obtain the Thevenin equivalent for the network show in the attachment between the terminals a and b.

    2. The attempt at a solution

    First I considered the voltage source alone (replaced current source by an open circuit).

    [tex] I = \frac{V}{Z_T} = \nfrac{20}{5+10+j5+j10+ (2*j4)} = 0.728A \angle-56.89\degrees [/tex]

    [tex] \therefore V_{AB}^1 = (0.728A\angle-56.89\degrees) * (5+10+j5+j4) = 12.74V\angle-25.92\degrees [/tex]

    Then I considered the current source alone (replaced the voltage source by a short):

    Current through upper branch = [tex] \frac { 2A\angle90\degrees (10 + j5 + j10 - j4 + (2*j4))||5}{10+j5+j10-j4+ (2*j4)} = 0.282A\angle29.83\degrees [/tex]

    [tex] \therefore V_{AB}^2 = (0.282\angle29.83\degrees)(j10 -j4 + j4) = 2.82V\angle119.83\degrees [/tex]

    [tex] V_{AB} = V_{AB}^1 + V_{AB}^2 = 10.529V\angle-17.25\degrees [/tex]

    Can someone please check this for me? I'm not sure whether I did some mistake and I need to know whether I understood these.

    Thank you for your time,


    Attached Files:

  2. jcsd
  3. Apr 20, 2010 #2

    The Electrician

    User Avatar
    Gold Member

    Your first calculation should be:

    [tex] I = \frac{V}{Z_T} = \frac{20}{5+10+j5+j10-j4+ (2*j4)}[/tex]

    Also, in this calculation, there is an error in your arithmetic:

    Current through upper branch = [tex] \frac { 2A\angle90\degrees (10 + j5 + j10 - j4 + (2*j4))||5}{10+j5+j10-j4+ (2*j4)} = 0.282A\angle29.83\degrees [/tex]

    I get:

    Current through upper branch = [tex] 2A\angle90\degrees \frac {(10 + j5 + j10 - j4 + (2*j4))||5}{10+j5+j10-j4+ (2*j4)} = 0.4131A\angle38.29\degrees [/tex]

    And, of course, these errors propagate into subsequent calculations.
    Last edited: Apr 20, 2010
  4. Apr 20, 2010 #3
    Ok, I agree with that part, found the mistake. However, do you agree with the rest of the working (especially the way I considered the mutual inductances)? I really wish to confirm whether I have the correct final answer or not. Now I got [tex]V_{AB}=9.225V\angle-13.79[/tex]
  5. Apr 21, 2010 #4

    The Electrician

    User Avatar
    Gold Member

    For the final answer, I get [tex]11.1156V\angle {-9.722} ^ \circ [/tex]

    Your handling of the mutual inductance seems essentially ok, but without showing all the calculations you showed in your first post, with the errors I pointed out corrected, I can't tell where you went wrong.
    Last edited: Apr 21, 2010
  6. Apr 22, 2010 #5
    I = \frac{V}{Z_T} = \frac {20} {5+10+j5+j10+ (2*j4)} = 0.728A \angle-56.89\degrees

    \therefore V_{AB}^1 = (0.728A\angle-56.89\degrees) * (5+10+j5+j4) = 12.74V\angle-25.92\degrees

    2A\angle90\degrees \frac {(10 + j5 + j10 - j4 + (2*j4))||5}{10+j5+j10-j4+ (2*j4)} = 0.4131A\angle38.29\degrees

    [tex] V_{AB}^2 = (0.4131A\angle38.29\degrees)(10 +j5 + j10 -j4 + 8j) = 8.87\angle100.53 [/tex]

    [tex] V_{AB} = V_{AB}^1 + V_{AB}^2 = 10.33V\angle17.76 [/tex]

    I can't see where I'm wrong...
    Last edited: Apr 22, 2010
  7. Apr 22, 2010 #6

    The Electrician

    User Avatar
    Gold Member

    Go have a look in post #2 where I said:

    "Your first calculation should be:"

    You're missing a -j4 term in the denominator of that V/Zt calculation.
  8. Apr 22, 2010 #7

    The Electrician

    User Avatar
    Gold Member

    I see another error. You should have:

    [tex] V_{AB}^2 = (0.4131A\angle38.29\degrees)(j10 -j4 + 4j) = 4.131\angle128.29 [/tex]
  9. Apr 22, 2010 #8
    You're right. Thanks a lot, I really appreciate you taking the time to go through my working.
  10. Apr 22, 2010 #9

    The Electrician

    User Avatar
    Gold Member

    You can also solve the system by writing a couple of loop equations. Replace the j2 current source in parallel with the 5 ohm resistor with a j10 voltage source in series with a 5 ohm resistor.

    Consider two loop currents, I1 and I2. I1 is a clockwise loop passing through the j10 voltage source, the 5 ohm and 10 ohm resistors, the j5 inductor, and the short across the a-b terminals. I2 is a clockwise loop going around the entire circuit.

    Then to solve for the two currents, solve this system. I1 will be the short circuit current through the shorted a-b terminals.

    [tex]\left[ \begin{array}{2}15+j5 & 15+j9\\15+j9 & 15+j19\end{array}\right]\left[ \begin{array}{1}I_1\\I_2\\\end{array}\right]=\left[ \begin{array}{1}j10\\-20+j10\\\end{array}\right][/tex]

    To find Zth, the Thevenin impedance, short the 20 volt source and the j10 source; connect a 1 volt source to the a-b terminals. The system then becomes:

    [tex]\left[ \begin{array}{2}15+j5 & 15+j9\\15+j9 & 15+j19\end{array}\right]\left[ \begin{array}{1}I_1\\I_2\\\end{array}\right]=\left[ \begin{array}{1}1\\0\\\end{array}\right][/tex]

    The solution to this system will give a current I1 whose reciprocal is numerically equal to the Thevenin impedance.

    Multiply the current I1 from the first system solution, by the Thevenin impedance, which is 1/I1 from the second system, and you will have Vth, the Thevenin voltage.

    This is how I solved your problem initially, and then examined your method to see where your errors were.

    It looks like the tex engine didn't properly format the 2x2 matrix on the left of the two systems above. Where you see a 4x1 matrix, convert it to a 2x2 by taking the terms row wise, two at a time.
    Last edited: Apr 22, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Thevenin Equivalent Circuit Date
Find the Thevenin Equivalent circuit Wednesday at 3:12 AM
Thevenin's equivalent circuit for capacitor Jan 28, 2018
Computation of Thevenin Equivalent Aug 7, 2017
Converting to a circuit's Thevenin equivalent? Feb 5, 2017