Very Simple Thevenin Question

Discussion in 'Homework Help' started by Hitman6267, Jun 2, 2010.

  1. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    When faced with a circuit like this, when calculating thevenin equivalents or anything. We can always say that the current ix is zero. Correct ?
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    When faced with a circuit where the current is shown going into a terminal which isn't connected to anything, then, yes, the current will be zero as long as the circuit topology remains as shown.

    But, to calculate the Thevenin equivalent, you may need to short the a-b terminals. Then the current Ix probably won't be zero.
     
  3. Eng_Bandar

    Active Member

    Feb 27, 2010
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    I think when you speak about thevenin almost find load between a and b terminals Why? because we want calculate max. power for load . Of couse, if nothing between a and b terminals Ix will be zero
     
  4. Ghar

    Active Member

    Mar 8, 2010
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    So... in summary no, it is not zero for Thevenin / Norton calculations.

    Short circuit current (ix) is not zero.
    Open circuit voltage has ix = 0 (since it's open circuit) and depends on the 10 mA source and V2 .

    Using those two you can get the Thevenin resistance as R = V / I
     
    Last edited: Jun 3, 2010
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    This circuit lends itself to a useful trick in finding a solution.

    Rather than shorting the terminals a & b you can apply a current drain of 10 mA.

    In this situation the left hand 10 mA source can be ignored in finding Vab.

    Which would yield

    Vab_10mA=100*(0.5-V2)/(R1+100)=(50-100V2)/(R1+100)

    We know Vth=Vab open cct which can be found from

    Vth=1-100(1+V2)/(R1+100)=(R1-100V2)/(R1+100)

    Hence

    Rth=(Vth-Vab_10mA)/10mA
    Rth={(R1-50)/(R1+100)}/10E-3
    Rth=100(R1-50)/(R1+100)

    If say R1=100 then Rth=100(100-50)/200=25Ω

    Interestingly, it would seem this circuit would have a negative Rth if R1 is less than 50Ω!
     
  6. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    That's what I mean't, I misphrased my question.

    Thank you for your replies.


    @t_n_K
    I'm a civil major, so all of that was wasted on me :p thank you though
     
    Last edited: Jun 4, 2010
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    No problem - it was probably of little interest most forum members, but sometimes there are ideas that may be worth posting just in case .....

    At least you have a general formula for Rth as a check on you work.
     
    Last edited: Jun 4, 2010
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    This can also be solved by traditional mesh and nodal methods.

    For the mesh method, replace the 10 mA current source and 100 Ω resistor with a 1 volt voltage source in series with a 100 Ω resistor. Now you have two meshes, the left one (I1 current) and the right one (I2 current) involving the a-b terminals.

    If the a-b terminals are shorted, we have:

    \left[ \begin{array}{2}100+R_1 & 50-R_1\\-R_1 & R_1-50\end{array}\right]\left[ \begin{array}{1}I_1\\I_2\\\end{array}\right]=\left[ \begin{array}{1}1+V2\\-V2\\\end{array}\right]

    Solving gives I_t_h=I_2=\frac{R_1-100*V2}{100*R1-5000}

    To find Rth, we can set all the internal sources to zero and apply a 1 volt source across the a-b terminals, giving:

    \left[ \begin{array}{2}100+R_1 & 50-R_1\\-R_1 & R_1-50\end{array}\right]\left[ \begin{array}{1}I_1\\I_2\\\end{array}\right]=\left[ \begin{array}{1}0\\1\\\end{array}\right]

    Solving, we get:

    I_2 = \frac{R1+100}{100*R1-5000}

    Since this is the current resulting from the application of 1 volt to a-b, Rth is the reciprocal of this value:

    R_t_h = \frac{100*R1-5000}{R1+100}

    Since Vth = Rth*Ith, we have:

    V_t_h =\frac{100*R1-5000}{R1+100}*\frac{R_1-100*V2}{100*R1-5000}=\frac{R_1-100*V2}{R1+100}


    Or, we could use the nodal method. There is only one node, so to calculate the open circuit voltage, Vth, we would have:

    \left[ \begin{array}{1}\frac{1}{100}+\frac{1}{R_1}\\\end{array}\right]\left[ \begin{array}{1}V_a_b\\\end{array}\right]=\left[ \begin{array}{1}\frac{1}{100}-\frac{V2}{R_1}\\\end{array}\right]

    Solving, we get:

    V_t_h = V_a_b = \frac{R_1-100*V2}{R_1+100}

    To get Rth, we zero all internal sources and apply a 1 amp current source to the a-b terminals:

    \left[ \begin{array}{1}\frac{1}{100}+\frac{1}{R_1}\\\end{array}\right]\left[ \begin{array}{1}V_a_b\\\end{array}\right]=\left[ \begin{array}{1}1-\frac{50}{R_1}\\\end{array}\right]

    Solving, we get:

    R_t_h = V_a_b = \frac{100*R1-5000}{R1+100}
     
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