Thevenin Equivalent Circuit

Discussion in 'Homework Help' started by BaconMan8910, Feb 12, 2018.

  1. BaconMan8910

    Thread Starter New Member

    Feb 9, 2018
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    VTH.png
    This is the circuit that I'm working with.

    I need to find the Thevenin equivalent. That is, I need to find the Thevenin resistance (RTH) looking into the circuit from the load, and the Thevenin voltage (VTH) or the voltage across the load.

    So, to find the Thevenin equivalent, I need to start by removing the load. This makes the circuit into a simple parallel circuit with the following attributes:

    RT = (R1+R3)//(R2+R4) = 5.76 ohms
    IT = 12.5A
    I1//3 = 8A
    I2//4 = 4.5A

    Assuming that all looks correct, I just need to figure out how to find the voltage across the load. Since (R1+R3)//(R2+R4), they will have the same voltage across them (72V) and dividing that voltage between the resistors in each parallel branch is relatively easy. What I'm unsure of is how the load will affect this circuit. For the sake of argument, we can assume the load to be a multimeter or, more accurately, what a multimeter would see when connected to terminals Vb and Va. Would the load, then, be the voltage drop between R2 and R4 (if we assume that Va is the positive terminal)?

    I have a similar issue when trying to find RTH. To find RTH, I remove the voltage source and find RT looking through the circuit from the load. The load, effectively, becomes a current source with infinite resistance (an ideal multimeter) for the sake of measuring resistance. If we assume Va to be the positive terminal, the current immediately splits into R2 and R4. Looking at the flow of current, it seems like R2 then becomes in series with R1 and R4 with R3. Is this correct? And, if so, are these sets of series resistors in parallel with one another?

    Thanks in advance. I'm still new to this and am trying to get my head around how everything relates.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    6,220
    Whoa!!!!

    Where does this IT = 12.5 A come from?

    First off, what IS IT? No where on your diagram do you define what IT is. You need to clearly indicate which branch this is the current for and, just as importantly, what direction it is. Don't make readers (or graders) guess what you mean. Engineering isn't about guessing.

    Assuming (and sentences that start with that word are another bad sign) that what you mean is the current coming out of the 72 V source, then it would appear that you have made one of the all time classic mistakes and decided to throw whatever resistance you have handy (your 5.76 Ω) and whatever voltage you have handy (your 72 V) at Ohm's Law, since it has an R and a V in it, and come up with some current (because the other variable is an I) without even bothering to ask or indicate what or where that current is.

    Ohm's Law is very, very specific. It relates the value of a resistance to the voltage ACROSS THAT resistance and the current THROUGH THAT resistance.

    So to use Ohm's Law to find the current coming out of the battery, you need the voltage of the battery and the resistance AS SEEN BY the battery? Your 5.76 Ω is the resistance seen by the load with the battery turned off (reduced to 0 V). NOT the same thing!

    Now, leaving this aside, you then have

    I1//3 = 8A
    I2//4 = 4.5A

    What is this? What does // mean?

    Whatever it is, how are you claiming that

    (a current) // (a resistance) = (a current)

    And, again, you are using currents that you haven't defined, either where they are or what their polarity is.
     
  3. BaconMan8910

    Thread Starter New Member

    Feb 9, 2018
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    IT = I (current) total.

    // denotes two resistors or branches of resistors in parallel. I’ve seen it written as II and as //. So, (R1+R3)//(R2+R4) is the branch of R1 and R3 in series, in parallel with the branch of R2 and R4 in series.

    I made a mistake when labeling current across branches. So I1//3 should be I1+3.

    So, I used my total resistance (5.76 ohms) and my total voltage (72V) to get the total current (IT) of 12.5A. I then divided it across the branches of R1+R3 and R2+R4 to get I1+3 of 8A and I2+4 of 4.5A.

    There’s a positive sign at the top of the voltage source. The current is running from positive to negative. So, from the top of the circuit to the bottom.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    So the Va and Vb voltage is ?
     
  5. BaconMan8910

    Thread Starter New Member

    Feb 9, 2018
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    I got 24V for Vb and 18V for Va, making VTH = 6V.

    For RTH, I got 5 ohms using (R1//R3)+(R2//R4).
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yep, if your Vth is Vb - Va instead of Va-Vb


    Yep, you are right.
    But next time to avoid confusion try to use this "||" for parallel connection instead of this one"//".
     
  7. BaconMan8910

    Thread Starter New Member

    Feb 9, 2018
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    Will do. Thanks.
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    So where did this "3" come from?
     
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