Help with Thevenin problem

Discussion in 'Homework Help' started by eldadoh, Sep 5, 2014.

  1. eldadoh

    Thread Starter New Member

    May 2, 2014
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    Hello guys, I started the course 'intrudction to electricity 1 ' and I have problem with this Thevenin queation . I Dont really know from where to start,
    I need to find the current and the voltage on RL Resistor

    What can I do with the right side of the problem, can i convert it to something ?
    I try to solve it , but don't think my way is correct but if you guys can help me with this it would be great.
    can you suggest me how to start this problem?
    I also need to solve it with Norton and superposition .
    Thanks a lot .
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Your solution looks good but you have 0.5A current source and 2K resistor.
    V = 0.5A * 2K = 1KV and for 5V voltage source 2.5K we have
    I = 5V/2.5K = 2mA
     
  3. eldadoh

    Thread Starter New Member

    May 2, 2014
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    so why i get 3.667mA? :/
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    How did 0.5 A=500 mA=500000 uA current source become 500 uA?

    How did 1 A=1000 mA current source become 1 mA?
     
  5. eldadoh

    Thread Starter New Member

    May 2, 2014
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    My mistake . I forgot to add 'mA' to the current sources. so it 0.5mA and 1mA
     
  6. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Again.
    How did 0.5 A=500 mA become 0.5 mA?
    How did 1 A=1000 mA become 1 mA?
    Do you have attention deficit disorder? The problem gives you one set of numbers, you magically turn them into completely different set of numbers.
     
  7. eldadoh

    Thread Starter New Member

    May 2, 2014
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    well, now i got 1.169A on the RL And voltage of 109.609V , where is my mistake , why i didnt get it
     
  8. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Using Mesh Current Methode I found current through Rl to be 1.137 A. Using Ohm's Law and 0.1 kOhm value of Rl I found voltage across Rl to be 113.7 V.

    My simulation got the same numbers that you got.

    sim.jpg
     
    eldadoh likes this.
  9. eldadoh

    Thread Starter New Member

    May 2, 2014
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    Thank you for trying to help me , but I need to solve it with thevenin , just tell me one thing : I need to use here source transformation or not ?
     
  10. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Source transformation. It is up to you to use it or not use it. You, the person who sovle the problem, decide to use source transformation or not to use it. It is not required. You decide for yourself. Some people do better when all the sources are voltage sources, other people do better when all sources are current sources. If you prefer to solve problems that have only one type of source, then do source trasformation. That way the circuit is transformed into one that you understand better.
     
  11. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    You can find Rth real easily. Remove Rl. Replace all independent current sources with open circuits, so they are gone now. Replace all independent voltage sources with short circuits. Now you have 2k in series with 4k, together they are 6k. 10k in parallel with 2.5k, together they are 2k. The 6k and 2k are in parallel, so we have 1.5k. That is your Thevenin Equivalent Resistance, Rth=1.5 kOhm.

    All you need to do now is remove Rl and find the voltage where Rl used to be. Notice that this voltage is the voltage across 10 kOhm resistor. So. Find voltage across 10 kOhm resistor, this will be your Vth.
     
  12. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Ok. I just did the Vth calculations using Node Voltage Methode (solved for voltage across 10 kOhm resistor). I got the same values for voltage across Rl and current though Rl as I showed in my simulation.

    I did not do any source transformations.
     
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