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Two Very Basic Circuit Questions

  1. Jun 19, 2009 #1
    Hey guys, I'm just trying to fully understand all these Circuit diagrams and stuff. If you can help me with these two questions I'd appreciate it, I thought about them while reading through my book. Sorry about the quality of pictures, I drew them in paint.

    Question 1:
    If I have a resistor, capacitor, resistor, capacitor in series like this picture below, even though they alternate, could I still look at them as if the 2 resistors were next to each other and the 2 capacitors were next to each other?
    q1.jpg


    Question 2:
    If I added an extra wire to the first circuit, would it still yield the same results? I'm assuming it would but I just want to make sure.
    q2.jpg

    Thank's for your help!
     
  2. jcsd
  3. Jun 19, 2009 #2
    Let me pull a 'clippit' and say "It looks like you are trying to do two-port models."

    Q1: Technically the two devices are not the same, but if what you are trying to do is build a 2-port device that "acts" the same way as your R-C-R-C combination, then it seems that you are on the right track.

    That is, the R-C-R-C combination with the two wires hanging off the sides acts the same way as your equivalent 2R-C/2 system when all you do is look at the circuit from the outside.

    In general: You can move linear circuit elements on the same branch around when analyzing the circuit, since the current through them would be the same, so long as you don't care about the node voltages between the elements, and there are no connections that come in between the elements.

    Q2: Here we have a problem. The circuit on the left acts like a resistor with series resistance 2R. The circuit on the right acts like a short circuit. To see why this is the case, try replacing that added wire with a resistor, say [tex]\tilde{R}[/tex], and let [tex]\tilde{R} \rightarrow 0[/tex].

    -------------
    Actually Q1 brings up an interesting dilemma. If you get two nonlinear devices, like diodes, and hook them up in series, would the circuit operate the same way (2-port wise) as if you interchange them, keeping the polarity constant? It seems that from a first analysis using a simple model it would, but putting in all the nuances in the IV curve, something fishy might happen.

    That is, suppose two diodes are connected cathode-to-cathode like this -->|--|<--

    How would the most honest physical 2-port model change when you reverse them to be like this --|<-->|--?
     
  4. Jun 20, 2009 #3
    Q1: yes, because all the components see the same instantaneous current.

    Q2; no, you've shorted it out! It wasn't a short before you added that wire.

    zprog: Consider that any non-linear series components will still always see the same instantaneous current, so their position in the series circuit is moot.
     
  5. Jun 20, 2009 #4
    I didnt even think about it being shorted out, but you're right. Thanks for your help both of you!
     
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