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Homework Help: Circuits: find current in the resistors

  1. Mar 22, 2004 #1
    Ok, so what i got here is a drawing of a circuit. It's rectangular, with points a & b making up one side of the rectangle. It's a symetrical diagram-use your imagination...

    4 __w__ 9
    __w__| |__w__
    | |__w__| |
    | 18 |
    | |
    a b

    The "w"'s are the resistors, and the number closest is it's resistance in ohms.

    a) Find the equivalent resistance btw points a & b.

    Easy. Add the resistance in parallel (1/R), then in series. I got 18.04 ohms.

    b) If a potential difference of 34.0 V is applied between points a & b calculate the current in each resistor.

    My main problem is I'm confused about the concept of "voltage drop". I know that the current is constant in series, but the voltage is not. That means by the time the current gets to the two resistors in parallel, the voltage is less that 34 V, right? How much less. Do you subtract the resistance of the first resistor (4 ohms) from the voltage-so the voltage entering the resistors in parallel is 30 V? Now voltage is the same in parallel- but also is the sum of the resistors? Hmmm, unless 25=30, I suspect an error in my thinking here...

  2. jcsd
  3. Mar 22, 2004 #2
    whoa that diagram got a little funky...basically point a connects to a 4 ohm resistor in series, which leads to a (junction?) errr two resistors in parallel: the top is 7 ohms, the bottom is 18 ohms, which connects to the final resistor in series at 9 ohms then to point b.
  4. Mar 22, 2004 #3
    Use the [ code ] tag next time: :smile:
    Code (Text):

       4   __w__   9
     __w__|     |__w__
    |     |__w__|     |
    |       18        |
    |                 |
    a                 b
  5. Mar 22, 2004 #4
    Just imagine you have a battery of 30V between points A and B. First you need to find the total current of the circuit, which is given by V/R(total). This is the current that goes through both the 4ohm and 9ohm resistors, so that's part of the question done.

    For parallel resistors, you know that the voltage on them is equivalent (and not the current). Therefore:
    [tex]V_1 = V_2 = I_1R_1 = I_2R_2[/tex]
    [tex]\frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{7}{18}[/tex]
    This is one equation with two unknowns, so you can't solve it. Luckily you also know that the sum of the two currents should be equal to the total current in the circuit, i.e:
    [tex]I_1 + I_2 = I_T = \frac{V}{R_T} = \frac{34}{18.04} = 1.885[/tex]
    So now you have 2 equations with 2 unknowns, easy to solve. :smile:
  6. Mar 22, 2004 #5
    thank you!
  7. Mar 22, 2004 #6
    So you don't "lose" any current anywhere in the circuit, right? It doesn't dissipate somewhere in the resistors or to an outside source?
  8. Mar 22, 2004 #7
    I'm not familiar with the term "losing current", sorry. Current is defined as the flow of charges (namely, electrons) in the circuit and those are certainly not going anywhere. The only thing you do "lose" is energy, which goes into heating the resistors as the charges flow through them.
  9. Mar 22, 2004 #8
    Mabye you meant to ask "is the current divided somewhere in the circuit"? To that the answer is yes. When two resistors are connected in parallel the potential drop on them is equal, but the current isn't, it's divided between the resistors like I showed above. :smile:
  10. Mar 22, 2004 #9
    Got it. Thanks again! :smile:
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