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Thevenins equivalent problem

  1. Apr 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve the thevenins equivalent of the circuit
    Thevinin.jpg

    2. Relevant equations


    3. The attempt at a solution

    I think this is wrong but I got 750 ohms for my Rthevinin and 2.3V for my Vthevinin. I checked using KVL with the load resistor set to 500 ohms and get a current of 0.0037A through the 500 ohm resistor but when I use the thevinins circuit I get a current of 0.0018A

    This is how I got Rth,
    Short the Voltage source, so the 750 is in parallel with the 1500 which is equivalent to 500 ohm resistor in series with the 1000 ohm to give a 1500 ohm in parallel with the second 1500 ohm to give my Rth of 750 ohms

    This is how I got Vth,
    Replacing the voltage source I see that the 1500(one connected in the separate branch) and 1000 ohm resistors have no current flowing through them. That leaves just the 1500 and 750 in series with the voltage source. So the voltage across the terminals would equal to the voltage in the 750 ohm resistor which works out to be 2.3V = Vth

    Can someone please tell me what I'm doing wrong ?, thanks
     
  2. jcsd
  3. Apr 4, 2015 #2

    gneill

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    Staff: Mentor

    Your Rth method and result look fine, but your Vth method and result are not: There will be a current through the upper 1500 Ω and the 1000 Ω resistors.
     
  4. Apr 4, 2015 #3
    If the current is flowing through all the resistors then is the voltage at the terminals measuring the voltage across the 750 ohm resistor only ?, so do I have to find the current flowing through the 750 ohm resistor and from their the voltage ?
     
  5. Apr 4, 2015 #4

    gneill

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    Staff: Mentor

    The voltage at the Out terminal will be the sum of the voltages across the 750 and 1000 Ohm resistors.
     
  6. Apr 4, 2015 #5
    I still don't think i'm getting the correct answer

    I worked out the equivalent resistance by taking the 750 in parallel with the 1000 ohm resistor and added that to the 1500 ohm in series to get 1928.57 ohms which is in parallel with the 1500 ohm (branched one) to get an equivalent resistance of 843.75 ohms. I then worked out the total current to be 0.0083A(7/843.75).

    I then used the current divider rules to work out the current in the 1000 ohm and 750 ohm resistors which I got to be 0.0021A(750 ohm) and 0.0015A(1000 ohm). I then worked out the voltages using ohms law to get 1.575 and 1.5 volts. Summing them I get Vth = 3.075V

    Don't think i'm doing this correctly, need some help, thanks
     
  7. Apr 4, 2015 #6

    gneill

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    Staff: Mentor

    I don't see how the 750 and 1000 Ohms resistors can be taken to be parallel.

    Let's rearrange the circuit diagram a bit to (hopefully) make things easier to see:

    Fig1.gif
     
  8. Apr 4, 2015 #7
    If I try it this way, then working out the equivalent resistance

    1500 + 1000 = 2500 ohms, which is in parallel to the 1500 ohm resistor. Working the parallel combination I get 937.5 ohms + the 750 ohm in series I get Req = 1687.5 ohms. So the total current would be I = 7/1687.5 = 0.0041A. Using the current divider rule I get I2 and hence the current in the 1000 ohm resistor to be 0.0041*1500/2500 = 0.0025A and the current in the 750 ohm resistor would equal the total current which is 0.0041A. So the voltage in the 1000 ohm resistor = 0.0025*1000 = 2.49V and the voltage in the 750 ohm resistor = 0.0041*750 = 3.075V. Adding the two I get Vth = 5.57V

    Not sure if that's correct but it still doesn't agree with the current I got using KVL. I know Vth should be 4.63V but I'm not sure how
     
  9. Apr 4, 2015 #8
     
  10. Apr 4, 2015 #9

    gneill

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    Staff: Mentor

    You want to check the details of how to apply the current divider rule.
     
  11. Apr 4, 2015 #10
    The easiest way to solve this is by superposition.

    Create the first Thevenin equivalent from the 1500 and 750 resistors in lower left sub-circuit.
    Create the second Thevenin equiv from the 1500 Ohm resistor at the top.
    Now by superposition, connect the two Thevenin equivalents to the Output terminal.
    From this simple circuit it is easy to calculate the Thevenin voltage and resistance.
     
  12. Apr 4, 2015 #11
    Thanks, I will look into that method
     
  13. Apr 4, 2015 #12
    Oh yes, sorry mistake that should be 0.0041*1500/4000. I now get what should be the correct answer of Vth = 4.61V

    Thanks gneill :smile:
     
  14. Apr 4, 2015 #13
    Looks good to me.
     
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