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Resistor & Ideal source network

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data
    http://img138.imageshack.us/img138/6717/hwprob.jpg [Broken]

    2. Relevant equations

    General intuition, KCL, KVL

    3. The attempt at a solution

    I've worked this out multiple ways to no avail. I think the easiest way to go about this is to set the left node to ground (0 potential) and go from there.
    EDIT: Using the assumption that my first statement below is correct, I found the solution of -46V. My second question still holds though.

    Here are a few things that none of my professors or TAs have been clear on:
    1) When you have current through a resistor (such as Io), using Ohm's law you then know the voltage. Something I'm very unclear on is which way the voltage drops across the resistor. In my class, we use current from positive -> negative as dissipated power with a positive sign (like in a resistor), and negative to positive as a source, with a negative sign. Does this mean that the potential at the bottom node is -2V?

    2) How can you have a potential across an ideal current source when the component doesn't have a resistance?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 14, 2012 #2
    1.) The voltage is always + sign where the current is entering the resistance and - sign where it is leaving the resistance. The voltage is across the resistor, so the voltage at the bottom node is -2V (relative to ground) only if the voltage at the top node is 0V (relative to ground). If the + sign is at node A and the - sign is at node B, it is important that all that means is V_AB = V_A - V_B. It does not mean V_AB > 0. It can still be negative.

    2.) An ideal current source has infinite resistance The idea is no matter what voltage is applied to the current source, you want a certain current to flow. Think about it this way: P = IV. If you have a current source, it is forcing current to be a certain value. Given the circuit, to do this enforcement, it may have to add power or subtract power. E.g. if it is a simple current source resistor combo, you obviously add power. But in other circuits, it may be that without the current source, you would have more current flowing through that branch, meaning the current source sucks power away to enforce the smaller current. So we can agree a current source contributes power. Well, if P = IV, we know there has to be an I through it and a V across it. Otherwise, no power is contributed.
     
    Last edited by a moderator: May 5, 2017
  4. Apr 14, 2012 #3
    Thanks! That's pretty helpful. So, for the first answer you're saying that since I arbitrarily chose a reference node, the -2V potential only means the relative difference between the potentials of the two nodes and not necessarily it's actual value?

    One more question regarding potential across a current source: There was another question where I needed to make a circuit so there was 0 potential across a current source so it wouldn't contribute to the overall power. Even with a 0 potential across the source, isn't the 3A still flowing since that's the very definition of an ideal source? How can you have current that doesn't contribute to the power of a system?
     
  5. Apr 14, 2012 #4
    For your first question, all voltages are always relative to two points. Even in circuits where ground is drawn, that is just a point arbitrarily chosen to use as a reference for all other voltages in the circuit. In other words, voltage across a resistor is ALWAYS in terms of a relative difference in potentials of the two nodes.

    For your second question, consider a 1V voltage source hooked up in series with a 1 ohm resistor hooked up in series with a 1 A current source. In this situation, the voltage source is already delivering the 1A, so it delivers all the power. The current source is sitting there doing nothing. It wants there to be 1 A and there already is. It brings in or sucks out zero power. The voltage drop across it will be 0V to reflect this lack of power delivery or consumption.
     
    Last edited: Apr 14, 2012
  6. Apr 14, 2012 #5
    Oh! So what you're saying is that we know it will always be 3A through that branch, according to the ideal source. However, the power supplied is a matter of how much of the 3A of current is actually supplied by this source. If all of the 3A is externally supplied, the potential across the current source will be 0 and thus also its power contribution according to P=VI. If less than 3A is already supplied, it will act as a power source; and if more than 3A is supplied it will dissipate the surplus power. In both of these latter cases will have a non-zero potential. Is this correct?
     
    Last edited: Apr 14, 2012
  7. Apr 14, 2012 #6
    Yeah.
     
  8. Apr 14, 2012 #7
    That's pretty fantastic, thanks so much for your help.
     
  9. Jul 29, 2012 #8
    what is the step function and impulse function of capacitor and inductor? please reply.....it s really urgent
     
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