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Non-linear device problem

  1. Jun 25, 2010 #1
    1. The problem statement, all variables and given/known data
    See figure

    2. Relevant equations

    3. The attempt at a solution

    Again, see figure I tried something and it looked promising but the answer doesn't work out.

    Any ideas?

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  3. Jun 25, 2010 #2


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    Could you explain your reasoning to get the equation

    [tex]\frac{1+v_Q/i_Q}{2+v_Q/i_Q} = 10[/tex]

    I don't see how you're applying Kirchoff's voltage law to get that.
  4. Jun 25, 2010 #3
    You know the 1 ohm resistor is in parallel with the 10V source, therefore the voltage across it is 10V. Now we can subtract from the 10V the drop across the other 1 ohm resistor which is 1*Iq, and subtract Vq. 10 - 1*Iq - Vq=0, we can plug in for Iq which is Vq^2+2*Vq. Solve for Vq, you'll obtain a positive and negative value but only the positive makes sense because of our polarity.

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  5. Jun 26, 2010 #4


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    When you combine the circuit resistances, first by adding the two in series, and then by calculating the two in parallel, you get a value of resistance (keep in mind SI units).

    In your equations you say this resistance is equal to 10 volts by KVL.

    [tex]\frac{1+v_Q/i_Q}{2+v_Q/i_Q} = 10[/tex]

    Resistance does not equal voltage.
  6. Jun 26, 2010 #5
    Thanks for the responses. I must have been out of it at the time putting resistance equal to voltage lol.

    I understand now.

    [tex] V_{q} = 2V [/tex]
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