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Electric current flowing through a resistor

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider the electric circuit shown in the figure. Assume that V = 10.4 V, R1 = R2 = R3 = R4 = R5 = 3.00 Ohm. What is the electric current flowing through resistor R2? I couldn't upload the figure, but you can find it on my cramster.com post:
    http://www.cramster.com/answers-jul...tand-determine-current-individua_883937.aspx"


    2. The attempt at a solution

    I can get as far as determining the equivalent resistance, but I am stuck as to how to use this to find the current flowing through a single resistor (in this case, R2). I would appreciate it if someone could explain to me the "how" and "why" behind finding the answer. Any help is appreciated!!!!!!!!! :-)
     
    Last edited by a moderator: Apr 25, 2017
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  3. Jul 11, 2010 #2

    mgb_phys

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    Firstly you need the total current flowing in the circuit, so you need the combination of R2,R3,R5
    Then you need the relative proportion flowing through R2 and the R3&R5 branch
     
  4. Jul 11, 2010 #3
    Ok, so I have found the equivalent resistance to be 8ohm, and the current across the equivalent resistance to be I=V/R=10.4/8=1.3 A. But, I am stuck as to what to do next. I don't know why, but I feel weak as to the theory behind individual resistors and current, voltage, etc across them. :-(
     
  5. Jul 11, 2010 #4

    mgb_phys

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    Remember the current through R2 is the same as through R5 - so all you are looking for is the current in the R2,R5 branch, which is just the ratio of the resistance in the two branches.
     
  6. Jul 11, 2010 #5
    Ok, so R2 and R5 are in series, so their equivalent resistance is 3ohm+3ohm=6 ohm. So, does that mean the current flowing through these 2 resistors is I=V/R, or I=10.4/6=1.73 A? And because they are in series and because it is the same current through both, that means the current through R2 is 1.73A, right? I am not sure I am fully understanding...
     
  7. Jul 11, 2010 #6

    mgb_phys

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    Since all the same resistors are the same then the resistance of R2+5 is twice R3 so there is twice as much current through R3 as R2 (or R5)
    If you need the exact current you just need to work out the equivalent resistance of the two branches - so you the resistance between the two dots. This gives you the resistance between the dots and so the total current through the two branches.
    Then you know that 1 third goes through the 2R branch and 2/3 through the single R
     
  8. Jul 11, 2010 #7
    Ok, I dont mean to be a pain... but how do I work out the equivalent resistance between the two dots (i.e. within the two loops)? If you could explain with a little more detail, I would appreciate it. I don't expect the answer to the problem, just a more direct explanation of how to find the path to the answer. Thank you for your help so far :-)
     
  9. Jul 11, 2010 #8

    mgb_phys

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    ok, lets call he resistance of each resistor 'R'
    The resistance between the two point is R and 2R in parallel (the two branches)
    So the total is = R*2R / (R+2R) = 2 R^2 /3R = R2/3

    This is then in series with R1 and R4 so the total resistance is R+R+R2/3 = 8R/3
    So the current I = V/R = V / 8R/3 = 3/8 V/R

    The current in R1 and R4 must be the same and the same as the current flowing between the two points.
    The current between these points is in the ratio of the resistances, so twice as much goes through R3 as through R2 and R5.

    So current in R2 is 1/3 of the current we just found = 1/3 * 3/8 V/R
     
  10. Jul 13, 2010 #9
    ok, thank you!!!!!!!!!!!!!!!!!!!!!!!! :-)
     
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