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Applying Thévenin's theorem

  1. Nov 9, 2015 #1
    1. The problem statement, all variables and given/known data
    TwMj2q3.png

    Find the current that flows through the ##8 \Omega##

    2. Relevant equations
    Thévenin's theorem
    3. The attempt at a solution
    the theorem says that I can replace all the circuit to a power source and a resistor connected in series.
    So first I need to connect all the power sources, Can I choose it to be ##20V-8V+4V=16V##?
    Secondly I need to fist the equivalent resistor, so first I will calculate the parallel ##4\Omega## and ##2\Omega## resulting a ##R_1=1.33\Omega## than to calculate ##R_1## with ##2\Omega## which are connected in parallel, resulting ##R_3=0.8\Omega## and now to connect all the others in series ##R_{Total}=0.8+4+4=8.8\Omega##

    So ##I=\frac{16}{8.8}=1.81A## is it right?
     
  2. jcsd
  3. Nov 9, 2015 #2
    There is an alternative: open the 8 ohms and calculate the voltage Vs; then short the 8 ohms and calculate the current Is
    Replace all of the circuit parts with a voltage source V and series resistor R=Vs/Io and now calculate the 8 ohm current as V/(R+8) .
    I have always found this method better than going around shorting voltage sources and stuff; and it directly yields the Thevenin equivalent circuit.
    If you have uncertainty about calculating the open/short circuit issues then you can apply the equivalent ( :) ) reasoning down to single component levels.
    For instantance starting from the left labeling resistors succesively we have and proceeding we can calculate the to node thevenin's:
    20*4/6 Vo and Is 20/2 which gives an R of 8/6 : ie. Thevenin of 13 1/3 volts and 1 2/3 ohms
    Now add R3 which doesn't change the voltage but does increase the resistance.
    Now add the 8V to the source.
    Then calculate the short circuit current when you add the parrellel R4 2 ohms
    Now we are cooking: and the final 4 ohms in series and the final voltage
    And you have built the equivalent circuit and have no doubt about the answer.
    **Of course you should learn and try the open/shorting of sources according to the class prescription (you do want to pass)**
    But I have _always_ found the constructive approach to be better in terms of certainty in the result.
    Incidently there are (at least) two other ways to do the calculation and yield a lot more information but the Thevenin approach works a lot of the time but don't forget the dual Norton approach, current source and parrallel resistance. The choice typically depends upon the type of load resistor/impedance (in your case 8 ohms) and it's variability.
     
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