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Homework Help: Thevenin Equivalent Question

  1. Jun 17, 2010 #1
    1. The problem statement, all variables and given/known data
    See Figure

    2. Relevant equations

    3. The attempt at a solution

    See figure.

    Did I do this right?

    I just want to make sure I'm understanding things correctly.

    Attached Files:

  2. jcsd
  3. Jun 17, 2010 #2
    No. You calculated the wrong value for I-thevenin (and thus for R-th as well). I'm not following what you're doing to find it, so I can't pinpoint the mistake. I will say though, from my own experience, current sources trick people up often. I'd stick with the voltage source and just write a 2 loop mesh analysis with I2 being Ith.

    By the way, in a simple circuit like this one with no dependent sources, you can short all sources and just find the equivalent resistance between the two points in question to find Rth.
  4. Jun 17, 2010 #3
    Is this what you mean?

    Attached Files:

    • Req.jpg
      File size:
      18.8 KB
  5. Jun 17, 2010 #4
    Yes, except you're not calculating the equivalent resistance correctly. That 15 ohm resistor only has 1 node in common with the 10 ohm resistor, so there is no way you can say they are parallel. Any parallel components share two nodes with each other whereas any series components share 1 node with each other.
  6. Jun 17, 2010 #5
    What am I doing wrong?

    Sorry if your help seems somewhat "hopeless" but I am indeed trying to understand this.

    EDIT: Doh, 5//10 + 15 = 18.3,

    Now the only part I have left to learn is the mesh analysis. If you don't have time to explain, I'll try googling and see what I can come up with.

    Thanks again!
  7. Jun 17, 2010 #6
    yes :) You may want to run through your Ith calculation again until you get the right value. You'll know now, because by using it, you'll calculate the same Vth as you just did.
  8. Jun 17, 2010 #7
    Well wait...

    If I'm looking for my Ith and I know my Vth and Rth, can't I simply apply Ohm's Law?

    Ith = Vth/Rth = 10/18.3 = 0.546A

  9. Jun 17, 2010 #8
    Yes. But doesn't it bother you that you incorrectly did the circuit analysis?

    I'll explain some mesh analysis for you. wait.

    First off, mesh analysis is useful for when you have voltage sources and resistors only. You can use it when there is a current source with a supermesh, but who cares about that for now. It uses Kirchoff's voltage law by saying "the voltage around a loop equals zero." Since you've already calculated Ith, I will use that circuit to teach you mesh (since i can't be doing hw for a poster if he already did it!).
    We'll start by writing the KVL for each loop:
    [tex] -V_s + V_{R_5} + V_{R_{10}} = 0[/tex]
    [tex] V_{R_{10}} + V_{R_{15}} = 0[/tex]
    We wrote these voltages with each loop having a defined current moving in a clockwise direction. As you should know, a resistor has + where the current goes in and a - where the current goes out. When writing the KVL for a mesh, just move along in a clockwise direction and write down the voltage with the sign of that which you approach. So you see the voltage source, we approached the negative sign, so I wrote -Vs. Also notice that the same 10 ohm resistor has current flowing into it in two different directions for what I wrote to be true. In the first loop, the current is entering from the top and in the second loop, current is entering from the bottom. We can do that! We just say some I entering the top = I1 - I2. Therefore, if we reverse the direction of the current, making it enter from the bottom, we'd write -i = I2 - I1.

    So we then substitute for each voltage across the resistor using ohm's law (v = ir). The current in the first loop with the voltage source will be I_1 and the current in the second loop will be I_2:

    [tex] -V_s + I_1 R_5 + V_{R_{10}} = 0[/tex]
    [tex] V_{R_10} + I_2R_{15} = 0[/tex]

    So the ones that I changed are the simple ones. the 5 ohm resistor has some current called I1 flowing through it clockwise and the 15 ohm resistor has some current flowing through it clockwise. But what about the center current? Like I said earlier, due to KCL, I going in from the top = I1 - I2. I going in from the bottom, the negative of the I going in from the top, = I2 - I1.

    [tex] -15 + I_1 R_5 + R_{10}(I_1 - I_2) = 0[/tex]
    [tex] R_{10}(I_2 - I1) + I_2R_{15} = 0[/tex]

    So what happens if we chose the "wrong" direction of the current? What if the current is not truly moving clockwise? No problem! The current you calculate will just be negative. In fact, you could do mesh analysis with both currents going counterclockwise, both going clockwise, or 1 going clockwise and the other going counterclockwise. It just makes sense to do a uniform approach to keep consistency.

    Also, for every new loop you add you add a new current. A general rule is n loops means n equations with n unknowns.

    People always asked me how I wrote mesh loops so quickly, so I'll tell you since it seems to evade most people I know IRL. All you have to do is sum the total resistance in a loop and multiply it by the current of that loop minus the sum of any shared resistances multiplied by the current that is sharing the resistor plus any additional voltages (from voltage sources). Then you have your loop equation. The first one would be -15 + (5 + 10)i_1 - 10I_2 = 0 and the second would be -10i1 + (10 + 15)i2 = 0.
    Last edited: Jun 17, 2010
  10. Jun 18, 2010 #9
    Thank you for all your help and the explanation of mesh analysis xcvxcvvc!

    I've got a few questions.

    These seems to work flawlessly and is very easy to follow. However, the confusion that remains is the confusion within my intution. I've drawn a picture to show you where I've become confused.

    I think you may have explained this in your post, but it's still not clear to me how the signs on the 10ohm resistors(I put it in red in the 2nd mesh analysis) are not interchanged.

    Other than that everything makes sense!

    Thanks in advance!

    Attached Files:

  11. Jun 18, 2010 #10
    On the top of the resistor, we can do a KCL, which states that the sum of the current entering and leaving the node equals zero, i.e. we did not create or destroy current/charge. So we will say the current in that branch is going from top to bottom.

    We have I1 entering it (going clockwise) and I2 leaving it (going clockwise), and since we defined the current as going from up to down, we have some unknown current I_down leaving it. So we can KCL these:
    I_down + I2 = I1
    again, this is current leaving = current coming in.
    Solving for this I_down, we get I_down = I1 - I2.

    So what is the current going up then? We just need to multiply I_down by (-1) to get I_up. I_up = -I_down = -(I1 - I2) = I2 - I1

    So when you write the second loop, you redefine the direction of the current as going up the branch, and you are still, therefore, entering the + node of the resistor.

    If you really wanted to, you could keep I_down and then write -Idown * R_shared for the second loop. But... this would be -I_down = I_up = -(I1 - I2) = I2 - I1 . It's the same thing!
  12. Jun 18, 2010 #11
    When did we do this? Was this implicit when I labeled the polarities of my 10 ohm resistor?

    This is where I always get confused between polarities and direction of current.
  13. Jun 18, 2010 #12


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    Staff: Mentor

    Resistors don't have polarities. Voltages have polarities and currents have directions. Your figure and equations look okay to me. With mesh analysis, you just pick the loop current directions (usually clockwise as you have used), and write out the KVL equations.
  14. Jun 18, 2010 #13
    Doh! :)

    What cases do you have negative voltages/currents? I seem to always make simple sign errors.

    Should I be looking at the polarities of the voltage across the resistor and which terminal the current in entering?

    For example, if current is leaving a node and passing through a resistor and experiences a voltage drop (+ to -) then that voltage will be positive, if it experiences a voltage rise (- to +) that voltage will be negative. Current leaving a node is positive and current entering a node is negative.

    Is this really how we define positive and negatives in circuits or does it come down to how to define circuit aspects how your circuit?
    Last edited: Jun 18, 2010
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