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Homework Help: Finding the Current through Resistors - Kirchoff's Laws?

  1. Dec 7, 2012 #1
    1. The problem statement, all variables and given/known data

    In previous parts of this problem, I solved that the current travels counterclockwise, and that I_A = \frac{2{\cal{E}}}{R_{2}+2R_{1}} .

    2. Relevant equations

    V= IR
    R[series,total] = (1/R1 + 1/R2 + 1/R3 ... ) ^-1
    KCL: sum of currents at a junction point equals zero
    KVL: sum of voltages at any point equals zero

    3. The attempt at a solution

    I'm not entirely sure how to tell if I'm supposed to be using Kirchoff's Laws, or how for that matter. I think I can use KCL, so that the current that passes through R2 is the sum of the currents passing through the two "R1" resistors, which I will identify as R1a and R1b. Because R1a and R1b are in parallel, they should have the same voltage yet different currents. The voltage that goes through R1a and R1b is the sum of the two EMFs in their parallel connection: 2ε. (?)

    The current that goes through R2 is I=V/R. The total voltage is 2ε. The total resistance is: (1/R1 + 1/R1)^-1 + R2 = R1/2 + R2.

    So: I = 2ε / (R1/2 + R2)

    This can be simplified: I = 4ε / (R1 + 2R2).

    The answer in the back of the book: I = 2ε / (R1 + 2R2).

    So my answer is twice the correct current. I think this has to do with the EMF I calculated above, marked with a (?). In addition to the mistake I made, it would be nice if someone could also clarify how I can tell by looking at a problem if I need to use KVL, KCL... I'm not comfortable with these topics, so I'm not sure if I even applied KCL to this one.
    Last edited: Dec 7, 2012
  2. jcsd
  3. Dec 7, 2012 #2

    Simon Bridge

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    You can use Kirchoffs Laws if you want to.

    Note: there are two circuit diagrams on that jpeg. You appear to be talking about circuit B:
    You are trying to exploit the symmetry of the circuit?
    Looking at that symmetry - could the current in each of the R1's be different?

    A quick way to check your assumptions would be to work out the Thevinin equivalent for the part of the circuit without R2 ... from there it should be easy to work out the current in R2.
  4. Dec 7, 2012 #3
    Thanks for noticing that I meant circuit B. Resistors in parallel, like these, must have the same voltage yet do not necessarily have the same current. However, these resistors have the same resistance (R1), and they have the same voltage, which I believe to be 2ε. So there currents should also be the same, through V=IR. I didn't originally think I was making-use of the symmetry, but now that I type this out, I guess I was.

    The equivalent to the part of the circuit without R2 is (1/R1 + 1/R1)^(-1), or R1/2. The voltage to this part alone is a total of 2ε. The current through just this part alone should therefore be 4ε/R1.

    This "part alone", separated from R2, is in series with R2... so now they have the same current. The current in the R2 resistor would be 4ε/R1. The way I tried understanding your method was incorrect.
  5. Dec 7, 2012 #4

    Simon Bridge

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    It follows that you have miscalculated the Thevinin equivalent.
    Doesn't this part of the circuit have two voltage sources in parallel?
  6. Dec 7, 2012 #5
    I know that in series, voltage is split around and shared between resistors, whereas the current stays the same. In parallel, I'm thinking of it as analogous and opposite: current is split around, and voltage stays the same.

    So if in a series, you calculate a current, and all the resistors have the same total current; then I'd think in parallel, when I calculate a voltage (2 emf), they must all have 2 emf as well.
  7. Dec 7, 2012 #6

    Simon Bridge

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    I am asking about the voltage of the voltage sources being in parallel.
    The battery-symbol things.

    If you put two 1.5V batteries in parallel - what is the total voltage?
    How does this apply to finding the Thevinin voltage?
  8. Dec 8, 2012 #7
    I looked it up online, and read that the two batteries in parallel should have the same voltage then... so then my mistake was thinking that their "same" voltage was the sum of them (2emf), when they really just had the same voltage that was indicated (1 emf). I think I can memorize this by thinking, that because in series it wouldn't make sense to add the currents of different circuit elements for the "same" current passing through them, you can't do the same thing with the "same" voltage for the parallel circuit's elements.

    Please correct me if I'm thinking anything that will bite me in the future. Also, you began your response by saying that I could use Kirchoff's Laws if I wanted to. What is it about this problem that makes it solvable without applying them?
  9. Dec 9, 2012 #8

    Simon Bridge

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    two equal voltage batteries in parallel do. Think of it in terms of potential, Try to avoid having to memorize new things as much as you can.
    It is the simplicity of the circuit, - it is two batteries and a load. You already know the answer to it.
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