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Math Help - Thevnin Equivalents

  1. #1
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    Thevnin Equivalents

    So for each of the figures I want to find the Thevinin equivalent resistance and voltage. I also want to find the maximum power dissipated by a load resistor placed between points a and b on all three figures. Finally, I want to know what Vab is when a 7-olm resistor is placed between a and b. Thanks in advance for the help!


    So for figure i) I found Rth=[(1/5)+(1/5)]^{-1}Ω+10Ω=12.5Ω
    I found Eth=(5Ω)(5V)(5Ω+3Ω+2Ω)^{-1}=2.5V and hence I have the Thevinin equivalent circuit.

    The maximum power dissipated is Eth^2/(4*Rth)=(2.5V)^2/(4*12.5Ω)=0.125 W.

    And if there was a 7Ω resistor placed between a and b Vab=(7Ω)(2.5V)/(7Ω+12.5Ω)=0.897V

    Next, for figure ii)
    If you short circuit both of the voltage sources, you can move the 10Ω resitor to the side where the 3V source used to be, and hence you have 4 resistors in parallel. Thus,

    Rth=[(1/(2Ω+3Ω))+(1/5Ω)+(1/10Ω)+(1/5Ω)]^{-1}=10/7Ω

    Next, to find the Eth, we short circuit one source at a time and find the contribution of each source.
    Due to the 3V source, we have:
    Req=[(1/5Ω)+(1/12.5Ω)]^{-1}=3.57Ω

    Thus, I=3V/3.57Ω=0.84A and therefore Eth=(0.84A)(5Ω)=4.2 V.

    Next, due to 5 V source:

    You can reduce the circuit to a single series circuit and Eth=(2Ω)(5V)/(2Ω+2Ω+3Ω)=10/7 V

    and therefore, due to both sources Eth=10/7V+4.2V=5.63V. Hence, we have our Thevinin equivalent circuit.

    The max power dissipated is =(5.63V)^2/(4*10/7Ω)=5.54 W

    And if a 7Ω resistor was placed between ab we have Vab=(7Ω)(5.63V)/(10/7Ω+7Ω)=4.68V

    Next, for figure iii)
    When you short circuit the voltage supplys, you end up with 3 parallel resistances.

    Rth=[(1/(2Ω+3Ω))+(1/5Ω)+(1/(10Ω+12Ω))]^{-1}=2.245Ω

    Next, the Eth due to the 2V source:
    Eth=(12Ω)(2V)/(12Ω+10Ω+2.5Ω)=0.9796V

    Now, due to 5V source:
    Eth=(4.074Ω)(5V)/(4.074Ω+3Ω+2Ω)=2.245V

    and together Eth=2.245V+0.9796V=3.22V and we have our THevinin equivalent.

    Pmax=(3.22V)^2/(4*2.245Ω)=1.16W

    and Vab due to 7Ω resistor between ab is =(7Ω)(3.22V)/(2.245Ω+7Ω)=2.44V
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  2. #2
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    I'll just comment on your answer to the first problem. Your Thevenin resistance and open circuit voltage calculations are correct. The maximum power transfer occurs when the load resistance equals the Thevenin resistance, which is 12.5 ohms. The current in such a situation is then

    i=\dfrac{V_{\text{oc}}}{2R_{\text{th}}}=\dfrac{2.5  }{12.5+12.5}=0.1\;\text{A}.

    Hence, the power dissipated in the load resistor alone is

    P_{\text{load}}=i^{2}\,R_{\text{th}}=0.125W.

    Finally, placing a 7 ohm resistor from b to a would result in a voltage divider just as you have there.

    In short, your answer for (i) is completely correct, though you might want to check your significant figures for the voltage divider at the end.

    If your second two problems are equally careful, then you're probably fine.
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  3. #3
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    Thank you very much. Additionally, I have a midterm on wednesday, so anyone who has the time to look over my Rth and Eth values for figure ii and iii will be greatly appreciated. I struggle with this class and I need all the help I can get. I also found those two figures exceptionally more difficult
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  4. #4
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    Problem (ii) seems a bit like a trick question. The open circuit voltage has to be 3V, because a is electrically connected to the positive of the 3V supply, and b is electrically connected to the negative of the 3V supply. Similarly, when you calculate the Thevenin equivalent resistance, you replace voltage sources with a short. That means a is shorted directly to b. There will be no resistance (obviously, ideal case, ignoring the resistance in the wires). So, the Thevenin resistance is zero.
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  5. #5
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    Makes sense. Does that consequently makes Eth equal to zero? Additionally the max power would not be able to be calculated, along with the 7 olm resistance. Ps I figured out figure iii
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    The maximum power transfer is essentially limited practically by the 3V power supply. Vab is just 3V, right?

    What did you get for Voc on iii)?
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  7. #7
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    That makes sense. So for ii, Eth=3V, Rth=0, and the maximum power dissipated is 0 W since Rth is zero. Also, for iii) I found Rth= 6.12 and Eth=Vab=(1.10-2)V=-0.897V, which means the voltage supply will be directed towards b having a higher potential.
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  8. #8
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    Due to the 2V source: The voltage at B is indeed 2V, however, there is a contribution of voltage at a. Hence, Eth=(1.10V+0.9795V)-2V=79.6 mV.
    Last edited by profound; March 9th 2011 at 05:30 AM.
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    I'm pretty sure that your calculation is wrong. Can someone verify?
    I found Vab due to the 5V source to be: V1=5V[(1/22-olm)+(1/5-olm)]^{-1}/9.0741-olm=2.2449V

    and due to the 2V source: V2=-(2V)[(1/5-olm)+(1/5-olm)]^{-1}/25.6-olm=-0.20408V

    and hence Eth=V1+V2=2.2449-0.2048V=2.04V!!!!!!!!!
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  10. #10
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    I don't think either of your answers are correct. Here's my working.

    Use mesh current analysis to find the current through the 12 ohm resistor. The left mesh is mesh 1, the right mesh is mesh 2. I get the following equations:

    -5+2i_{1}+5(i_{1}-i_{2})+3i_{1}=0

    2+5(i_{2}-i_{1})+10i_{2}+12i_{2}=0, or

    10i_{1}-5i_{2}=5

    -5i_{1}+27i_{2}=-2.

    The solution to this system is i_{1}=0.51,\,i_{2}=0.0204. Therefore, the voltage across the 12 ohm resistor is given by

    V_{ab}=12i_{2}=12(0.0204)=0.2448\;\text{V}.

    Note that b is at a lower potential than a, due to the orientation of i_{2} and the passive sign convention. For the Thevenin resistance, I get

    12\|(10+5\|5).
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  11. #11
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    When R_{Th}=0 do we say the maximum power dissipated at a resistor through ab is infinity, or would we simply state undefined?
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  12. #12
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    When R_{\text{Th}}=0 do we say the maximum power dissipated at a resistor through ab is infinity, or would we simply state undefined?
    Neither. If the Thevenin resistance is zero, there are two things you can say about the system, and both are related to an actual physical setup.

    1. The Thevenin resistance is never actually zero unless you're dealing 100% with superconductors (not likely at this stage). You'll have the internal resistance of the wires which, though small, is not zero.

    2. As the answer to the question, I would probably say that the maximum power dissipated is determined by the power limitations of the voltage source in combination with the actual resistance of the load that you hang onto the Thevenin equivalent circuit. For example: suppose your voltage source is a 24V, 3A max power supply, and your load resistance is 1000 ohms. Thus, the maximum power output of the power supply is 24 x 3 = 72 watts. However, you may not be able to get that, because of the resistor value. Can you get 24V? Well, Ohm's Law states that if you had 24V across the resistor, the current would be 0.024 A, which is definitely possible with that power supply. On the other hand, suppose you wanted to get the maximum of 3A out of the power supply. Could you get that? Well, if you could, the voltage across the resistor would be 3000V, which is definitely not possible. What you have here is a maximization problem:

    maximize P = IV subject to V = 1000 I, 0 < V < 24, and 0 < I < 3. Those last three expressions really mean that 0 < I < 0.024. Since P = I^2 R = 1000 I^2, it is clear that the maximum power occurs when I is the maximum allowed. Hence, the maximum power is 1000 (0.024)^2 = 0.576 W.

    In deriving this value, you can see that both the physical limitations of the power supply and the actual value of the resistor were absolutely essential. That is why I say that the maximum power dissipated will depend on those two things.

    In addition, you can see that the value I got was neither infinite, nor was it undefined. There's a definite procedure for finding out, given the particular voltage source and load, what the maximum power is. If you don't know one or both of those things, however, then you can't say what the maximum power dissipated is.

    Incidentally, in the example, you could get 72 W dissipated, if the load resistance happened to be 8 ohms (24 V divided by 3 A) AND the resistor was rated for that many watts (such resistors do exist). If the resistor was not rated for that many watts, and you tried to get that much power dissipation, you'd end up with charcoal. I've done that. Not recommended.

    Hope this helps.
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  13. #13
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    You're such a good help! So for my example, since it doesn't state the what the load is, and all I have is a value for Eth, then I would merely state that there is not suffice information for determining the maximum power dissipated.
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  14. #14
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    Quote Originally Posted by quantoembryo View Post
    You're such a good help! So for my example, since it doesn't state the what the load is, and all I have is a value for Eth, then I would merely state that there is not suffice information for determining the maximum power dissipated.
    That sounds good to me.
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