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Current algebra

  1. Jan 8, 2016 #1
    1. The problem statement, all variables and given/known data
    An electric circuit consists of 3 identical resistors of resistance R connected to a cell of emf E and negligible internal resistance. What is the magnitude of the current in the cell? (in the diagram two of the resistors are in parallel with each other then the other in series with that pair)

    2. Relevant equations
    I = V/R

    1/R = 1/r + 1/r +...

    R = r1 + r2

    3. The attempt at a solution

    It's a multiple choice question and the answer is 2E/3R but I can't see where that has come from.

    If I said that R was 2 ohms then the combined resistance would be 3 ohms but that doesn't help. I can't see where the 2/3 has come from please help, thank you
  2. jcsd
  3. Jan 8, 2016 #2


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    Staff: Mentor

    Evaluate the total resistance symbolically.
  4. Jan 8, 2016 #3

    Can you calculate what the equivalent resistance of this circuit is?
  5. Jan 8, 2016 #4
    I think it's (R1xR2)/(R1+R2) for the parallel part then (R1xR2)/(R1+R2) + R to add the series part in but don't know how to combine those?
  6. Jan 8, 2016 #5
    In order to add fractions, you need to have the same denominator in both.

    and remember that all three resistors are identical. This will come in handy when simplifying your result.
  7. Jan 8, 2016 #6
    Ok so ((RxR)/(R+R)) + R

    (2R/2R) + R? I don't think that's right but don't understand how I would get the same denominator either
  8. Jan 8, 2016 #7
    Not quite.

    ##\frac{R \times R}{R + R} + R = \frac{R^2}{2R} + R ##

    If you were asked ##\frac{1}{2} + 1 = ?##, you would need to recognise that this is equivalent to ##\frac{1}{2} + \frac{2}{2} = \frac{1 + 2}{2}##.

    You need to use this very same trick.
  9. Jan 8, 2016 #8
    1/2 R2/R +R/R

    Do I have to separate the square now? Sorry, still can't see how I'm going to get the 2/3. Really appreciate your help so far
  10. Jan 8, 2016 #9
    (1/2R2 + R)/R?
  11. Jan 8, 2016 #10
    ##\frac{R^2}{2R} + R##

    If you simplify this expression, what do you get?

    Edit - Sorry, this is my final edit. I misread what you typed.
    Last edited: Jan 8, 2016
  12. Jan 8, 2016 #11
  13. Jan 8, 2016 #12
    If you cancel the ##R## terms in the fraction, you get ##\frac{R}{2} + R = \frac{3R}{2}##
  14. Jan 8, 2016 #13
    Thank you so much that makes sense!
  15. Jan 8, 2016 #14
    You're welcome.

    To recap (so you have the entire argument in one place):

    Two resistors in parallel: ##\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} ##.
    The equivalent resistance of this pair is therefore ##\frac{R_1 \times R_2}{R_1 + R_2}##

    Two resistors in series: ##R_{eq} = R_1 + R_2##

    In your situation, you treat the equivalent resistance of the parallel setup, as a single resistor in series with your third resistor.

    ##R_{total} = \frac{R_1 \times R_2}{R_1 + R_2} + R_3##

    Since ##R_1 = R_2 = R_3##

    ##R_{total} = \frac{R^2}{2R} + R = \frac{R}{2} + R = \frac{3R}{2}##

    and then it is just the case of applying ohm's law.
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