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Homework Help: Kirchoff's Law Exercise

  1. Sep 14, 2014 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    V= iR
    Kirchoff's law for voltage

    3. The attempt at a solution

    On the left-most part of the circuit , I did use KVL (voltage law) and I ended up with Vs = i0 (R1+R2) which basically boils down to i0(2R) since R1 = R2 =R3 ...

    I'm not even sure if both parts of the circuits are parallel or not... I'm kinda lost .
  2. jcsd
  3. Sep 14, 2014 #2


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    When you are asking folks for help, it's a good idea to tell them WHAT you are asking for help with.

    The 2nd side of the circuit has a controlled source in it. Did you not see that? I don't see how you can even ask if the two parts are in parallel. They happen to have a single wire between them, and that can be taken as a ground, so they are not effectively connected at all, just that one is dependent on the other due to the dependent source.
    Last edited by a moderator: Sep 14, 2014
  4. Sep 14, 2014 #3
    What do I do with the other part of the circuit? Does making R3 & R4 equivalent the same as taking the potential difference for one path only? (including the controlled source)

    Edit : How can it be ground when it connects both circuit together? Shouldn't a ground wire be attached to something else not from the circuit ?
    Last edited: Sep 14, 2014
  5. Sep 14, 2014 #4


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    Parallel means the voltage across them will be the same

    Edit : How can it be ground when it connects both circuit together? Shouldn't a ground wire be attached to something else not from the circuit ?[/QUOTE]
    It is traditional to take the bottom horizontal wire in a simple circuit to be ground even if it is not shown with a ground symbol on it but in this case it really doesn't matter. The wire between the two parts of the circuit can be removed as it has zero effect on the circuit and it irrelevant.
  6. Jun 10, 2015 #5
    It is worth noting that this circuit is a model of actual electronics. This model includes in it a current-dependent current source. The diamond element tells you that it is a dependent-source. The arrow within the diamond element tells you that it is a current source. And what is it dependent on? It is somehow dependent on I0. Thus, you have a current-dependent current source. Don't worry at this time about how this actually happens; that is not the point of the exercise. Just know that this is a model. What actually happens is that you have some complex electrical components (which most likely include nonlinear electrical components). The complexity of the actual circuit is probably very complex. But all you know (and all you need to know) is that a current in one part of the circuit is somehow dependent on the current in another part of the circuit? How is this? It doesn't matter. It is not magic, but as far as you are concerned, it might as well be magic. The point is, you don't need to know how this dependency was established.

    Okay, now note that the left side of the circuit and the right side of the circuit are connected by a single wire. According to KCL, what is the current in this wire between the two circuits?

    Also, know that when you are dealing with current sources, you can only find out the voltage across a current source, by finding out what the voltage is across some element that is parallel with that current source.

    I hope this helps.
  7. Jun 10, 2015 #6


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    Well it's been many months since this post, just in case the OP were to come back and look for his or her own interest (As it's impossible for this to still be course work..)

    The current through the first branch can be represented as $$I_0 = V_S/(R_1+R_2)$$ The current source draws $$\alpha I_0 = \alpha V_S / (R_1+R_2) $$ This current is drawn from both of the right most branches in parallel, so we can say that $$\alpha V_S / (R_1+R_2) = V_0\left(1/R_2 + 1/R_4\right) $$ Finally solving, $$V_0/V_S = \alpha R_3R_4 / (R_1 + R_2)(R_3 + R_4)$$ If all resistors are equal and I didn't make a mistake, alpha should be 40 to get a voltage ratio of 10. Just to echo what was said, the most important thing to take from this is that this is frequently how transistor circuits are analyzed to size components for a specific gain considering the "small-signal" model of the component.
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