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Resistance and Current-Direct Current Circuits

  1. Jul 22, 2010 #1
    Hi all, I'm new here, been havin' a good ol' tussle with this one. Has me frustrated because it seems pretty elementary.



    1. The problem statement, all variables and given/known data
    The circuit diagram has been drawn out in the attachment

    The problem asks what is the current through the 20 ohm resistor.
    2. Relevant equations

    Parallel: 1/Req=1/R1+1/R2+...1/RN
    Series: Req=R1+R2+...RN

    I=V/R
    3. The attempt at a solution

    I've added the 10 and 5 ohm resistors to make 10/3 ohm to simplify that area of the diagram.
    I've also added the 20 and 5 ohm resistors on the bottom end of the diagram to make 25 ohm.

    Thus, adding these two equivalent resistances gives me 85/3. This leaves the resistor to the left of the battery.
    Adding that one to this value as per the series rule of equivalent resistances gives me 115/3.

    Using this equivalent resistance, I can find the total current, which comes out to be

    I=25/(115/3)=75/115=.652 A.

    From here, I am at a loss as to how to use this value to calculate the specific current at the 20 ohm resistor.

    Am I even using the proper voltage to calculate the total current?

    The correct answer to this problem is 227 mA.

    Thanks, any help would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Jul 22, 2010 #2
    Hi glamdring22,

    I=V/R is a useful formula. It is not only applicable to V and I through the main circuit, but also applicable to V and I through any resistor. Let's try to exploit it :wink:

    You have done a great job to find I through main circuit. Now let's see what else we need. Here are some analyses:
    - I through the 20-ohm is also I through both 20-ohm and 5-ohm! (how easy to deal with resistors in series!)
    - In order to find I through them, we must know V through both 20-ohm and 5-ohm, as pointed out by the relation I=V/R, right? :wink:
    - That V is also: a) V through the remaining 5-ohm, b) V through the lower 10-ohm, and c) V through the upper 10-ohm and the power supply.
    Now which would you choose to obtain the solution? Make it fast! Don't forget the relation I=V/R :smile:

    No, you cannot add them like that :frown:
    You know, the (5-ohm + 20-ohm series) and the (10-ohm and 5-ohm in parallel) are parallel, so adding their resultant resistances directly is not right :smile:
     
    Last edited: Jul 22, 2010
  4. Jul 22, 2010 #3
    (Regarding your correction, for the (5 +20 in series)+(10+5 in parallel) I got a corrected answer of 50/17 ohms.)


    that leaves the 10 ohm resistor to the left of the power supply and the newly solved equivalent resistance. Am I correct in understanding that these two final resistors are in series? Thereby directly adding up?

    Going from there I have: (10 Ohm resistor the the left of battery)+{(5+20 in series)+(10+5 in parallel)}= 220/17 ohms.

    using this result, I calculated a corrected result for the main current:

    I=V/R=(25*17)/220=1.93 A.

    Ok, so with this main current I'm still not seeing what you're getting at.
    Do you mean that I should use the main current to solve for V in one of the scenarios that you have spelled out?
     
  5. Jul 22, 2010 #4
    Yes, they are in series, and they have to be added up.

    Good work! :approve:

    Yes, of course :smile:
    Okay. Now with the main current, you can solve for V through the upper 10-ohm resistor, can't you? (that's why I mention the I=V/R!) :wink:
    Then you will have: U (power supply) = (V through the upper 10-ohm resistor) + (V through the series 5-ohm+20-ohm) :smile:
     
  6. Jul 23, 2010 #5
    Oh lordy. I think I got the answer.

    While I am very relieved, I would very much like to clear something up, for future problems.

    I did what you said and did the operation

    25V(power supply)=19.3(voltage through the upper 10 ohm)+(voltage through 20+5 series)

    this resulted in a voltage of 5.7 V.

    Dividing this by the combined series resistance of 25 ohm gives me the correct answer of 227 mA.

    This is all fine and dandy, but the issue is, how would I know to do this?
    What is the theory behind the idea that I had to use the voltage specifically through the upper 10 ohm resistor?

    Does this mean that the 25 ohm combined series resistance carries the same current and voltage as the individual 20 ohm? I'm still fuzzy on this theory.
    Thanks, sorry to resurrect an old dead thread.
     
  7. Jul 24, 2010 #6
    Of course the current through the combined 25-ohm resistor is the same as the current through the 20-ohm one, because they are in SERIES.

    So the question is, how we know we should do this, not do that. Well, I would say, first, it's because we aim to find current through the 20-ohm resistor, and second, it's experience. This is how I thought:
    1 - Because I want to find I through the 20-ohm resistor, I may have to find V through it. But wait, trying to find V will leads me to nowhere, because I haven't connected something from the the rest of the circuit to my calculation! I have to get out and see something bigger, otherwise I will get myself stuck around with the 20-ohm resistor. And something bigger is the series (5-ohm + 20-ohm), and surprisingly, I through the series = I through the 20-ohm resistor! So I have to find I through the series.
    2 - Again, to find I through the series, I have to find V across the series. But I'm stuck again if I consider the series only. Again, I have to get out. But I have 3 paths to go: the upper series (10-ohm + power supply), the middle (10-ohm), and the lower (5-ohm). The middle and the lower ones seem to be a mystery to me, as I know neither their currents nor their potential differences! Meanwhile, I know current through the upper - at least, I have a clue here, so I choose it.
    3 - To connect the series (5-ohm + 20-ohm) and the upper series, I have one equation: V_(5-ohm + 20-ohm) + V_(upper 10-ohm) = U. Now I have current through the upper series, I can calculate V_(upper 10-ohm). I have all I need.
    Hope this help :smile:
     
  8. Jul 24, 2010 #7
    Interesting, thank you for being so patient with me.

    And what of the current through a parallel circuit? I am just really interested in knowing all the different possibilities.
     
  9. Jul 24, 2010 #8
    What do you think? Let's take it as an exercise :smile:
     
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