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A device connected to a current source and then resistor

  1. Feb 4, 2012 #1
    Forgive the title, I was unsure of how to briefly describe the problem.
    Attached are diagrams.

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

    The device shown in Figure P2.5 can be modeled as a current source in parallel with a resistance. When this device is connected to a 5[mA] current source, so that 5[mA] enters the device at terminal A and leaves at terminal B, then the device delivers 296[mW] to the 5[mA] current source.
    Then, when the same device is disconnected from the current source and is connected to a 10[V] voltage source, so that there is a 10[V] voltage at terminal B with respect to terminal A, then the device absorbs 16.5[mW].
    Find the power absorbed by the device when the device is connected to a 10[kΩ] resistor.

    2. Relevant equations
    v=ir, p=iv, KVL, KCL

    3. The attempt at a solution

    First with the Current Source [Circuit one]
    power.del.by.device = power.abs.by.5ma = .296[W]
    p=iv so...
    v(5ma current source) = 59.2 [V]

    KCL on top node...
    -.005 + (59.2/Rx) - ix = 0

    Now with the Voltage Source [Circuit two]
    power.abs.by.device = power.del.by.voltagesoruce = .0165 [W]
    p=iv so...
    i(voltage source) = .00165[A]

    KCL on top node...
    -.00165 + (10/Rx) - ix = 0

    Solve the linear equations...
    Rx = 14686.57 [Ω]
    ix = -.000969

    Third condition (resistor)
    I forgot to make a diagram for this one. The new resistor has a current flowing upward. Its poles are positive on top and negative below.

    power.abs.by.device = power.del.by.10kresistor
    p=i^2 (r)
    we need i at the resistor...

    KCL on top node...
    -ix - i(10k resistor) +i(middle resistor) = 0

    KVL on left loop
    -i(10k resistor) +Rx*i(middle resistor) = ...
    -i(10k resistor)*10000+14686.57*i(middle resistor) = ...
    -i(10k resistor)*10000+14686.57*(ix + i(10k resistor))= 0 (Solvable)

    i(10k resistor) = .003037[A]

    p= i^2 (r) = .003037^2 (10000)
    p.abs.by.dev = 92.2 [mW]

    My main concern is that their is something wrong with my understanding of current flow, when to flip signs etc. I know that in all my circuits I have the current iQ (which i refer to as the current across the introduced component) which goes the opposite direction of what it "should" be going (positive to negative). There are several questions like this on my HW which is another reason I want to be sure I get this one right. If something was left unclear please do ask.
     

    Attached Files:

  2. jcsd
  3. Feb 4, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    As you've discovered, you need to be a bit careful about the signs of things.

    Consider a simple circuit consisting of a current supply driving a resistor:

    attachment.php?attachmentid=43491&stc=1&d=1328390957.gif

    The resistor is dissipating power, so the current source must be producing it. Note that the potential that appears across the current source is positive. So the current source is delivering power at a rate p = IV. By extrapolation then, the current supply would dissipate power if that potential is negative.

    In your first circuit, where the current source is said to be absorbing power this implies that the potential across the current source should be negative. It's taking charge carriers from a higher potential at its bottom and delivering them to a lower potential at its top, and absorbing the energy involved in doing so.

    For the second case, where the voltage source is placed across the device, pay attention to the stated polarity. Which polarity is connected to terminal B?

    You might also want to consider that if the device can be modeled as a current source in parallel with a resistance, then via Norton and Thevenin, it can also be modeled by a voltage source in series with a resistance. This can be handy if you're more comfortable dealing with series circuits.

    Another hint: The problem does not state that the "resistance" in the device model is in fact a typical resistor.
     

    Attached Files:

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