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V-i characteristics for voltage divider with current source

  1. Mar 9, 2015 #1
    This is a problem from Foundations of Analog and Digital Electronic Circuits by Agarwal & Lang, I'm going through it by myself, I'm not in a course, so I'm depending on you friendly forum people for help! I really like this book but it has a lot of errors, perhaps this is another error.

    1. The problem statement, all variables and given/known data


    Sketch the i-v characteristics for the network. Label intercepts and slopes.
    2_8_e.jpg

    2. Relevant equations
    The solution PDF for this textbook says that the answer is
    i = (v/4) + (v/5 + 2)
    i = (9/20)v + 2
    So the slope is 9/20 and the i-intercept is 2 Amps.

    3. The attempt at a solution
    My answer is i = v/4 + 2, which I got from connecting a current source to the terminal and solving the circuit for i. So the slope is 1/4, the i-intercept is 2 Amps.

    I tried simulating the circuit on partsim.com with a DC sweep and I think it confirmed my answer, but I don't really know my way around circuit simulators yet so maybe I messed up.

    I understand that the solution PDF got its answer by plugging Ohm's law into i = i1 + i2, where i1 is current through the 4 Ohm resistor and i2 is current through the 5 Ohm resistor.

    But didn't it get i2 wrong? Shouldn't i2 = 2 Amps, because the current through the 5 Ohm resistor goes directly into the ideal current source which is set at 2 Amps?

    So in my view, the 5 Ohm resistor could be any resistance, it doesn't matter, because there will always be 2 Amps going through it. Am I correct? (Obviously for real life elements this wouldn't hold up but I'm talking ideal elements here). I still am not totally comfortable with the intuition behind ideal current sources so confirming/denying this would really help me out.

    Thanks!
     
  2. jcsd
  3. Mar 9, 2015 #2

    gneill

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    Your answer looks good to me.
     
  4. Mar 10, 2015 #3

    CWatters

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    I agree. I looked at two cases..

    Vin = 0 so no current flows through the 4 Ohm. Means Iin = 2A
    Iin = 0 so all of the 2 Amp flows through the 4 Ohm. Means Vin = -8V.

    Plot both points. Slope is 2/8 = 1/4.
     
  5. Mar 10, 2015 #4

    rude man

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    My answer is i = v/9 + 10/9
    Solved by summing currents to 0 at the output (top of the 5 ohm), & solving for i (and output voltage).
    EDIT I drew the picture wrong. :H
    Post #5 is correct. Interesting that the 5 ohm plays no part in the i-V characteristic.
     
    Last edited: Mar 11, 2015
  6. Mar 11, 2015 #5

    CWatters

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    The 5 Ohm is horizontal. Do you mean the top of the 4 Ohm? I tried applying KCL to the node at top of the 4Ohm and get...

    Let into the node be +ve then...

    Iin - I4ohm - 2 = 0......................(1)

    Apply Ohms law to the 4Ohm..

    I4ohm = Vin/4 ...........................(2)

    substitute (2) into (1)

    Iin - (Vin/4) - 2 = 0

    Rearrange..

    Iin = (Vin/4) + 2

    Same result as my previous attempt. Slope = 1/4, intersects at +2A
     
  7. Mar 13, 2015 #6

    NascentOxygen

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    This is wrong. It should be i = (v/4) + 2

    The current through the 5Ω resistor is a fixed 2A. End of story.
     
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