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Potential across a Resistor for an Ideal Battery

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data

    An ideal battery with an emf of = 14.0 V is connected to a network of resistances with R1 = 6.00 , R2 = 12.0 , R3 = 4.00 , R4 = 3.00 and R5 = 5.00 . What is the potential difference across R5?

    __R1__
    __________R3__
    __R2__

    __R4_____R5__

    __emf battery__


    2. Relevant equations

    EMF = iR

    Req = R + R + R in series, 1/Req = 1/R + 1/R + 1/R in parallel

    i=V/R


    3. The attempt at a solution

    I have found that R1 and R2 are in parallel, which R12 is then in series with R3. R4 and R5 are in series with each other, and thus R123 and R45 are in parallel.

    Finding the equivalent resistance, 1/R1 + 1/R2 = 1/R12 = 4, then R12 + R3 = 8. R4 + R5 = 8,

    And 1/R123 + 1/R45 = 4 for the equivalent resistance.

    Now that I have formed this circuit into a single loop with Req = 4, and EMF = 12, I can find the current, i, running through it, thus 12/4 = 3 = i

    The question asks for the potential running across R5. I know that the current will be the same everywhere in the loop, so I solved for the lower branch with R4 and R5.

    Knowing i=3, and Potential is V=i(r), I get 3*5 to be 15, but this is not correct.

    The reason I'm trying to figure this out is because I have the feeling it will be on the next test, and I can't seem to get it.
     
  2. jcsd
  3. Sep 23, 2011 #2

    PeterO

    User Avatar
    Homework Helper

    Couple of things:
    Your original statement said emf was 14, but you used 12 later on??

    Secondly, with the two parallel branches, the current through the battery is split between the two branches, so the full current does not pass through R5.

    Secondly, you don't need to find the total resistance to solve this - but maybe that was a separate part of the question.
    It sounds like R5 was part of a branch which followed Battery - R4 - R5 - back to battery, so the emf was effectively, merely applied to those two resistors
     
  4. Sep 23, 2011 #3
    http://i.imgur.com/WhOXp.png

    Here is a better view of the circuit.

    Sorry, it is 12 V, not 14 V.

    I thought at first that I could make a small loop just across R4 and R5, but that didn't work out, probably because it has to be wired through both since they are essentially connected in a parallel circuit.

    Ignore the values posted for this problem, just the circuit is the same. I'm trying to solve the problem in the back of the book first, then applying that concept to the homework.
     
  5. Sep 23, 2011 #4

    PeterO

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    Homework Helper

    Sorry to disagree with you, but to remove R1, R2 and R3 from the circuit will make absolutely no difference!! Perhaps it didn't work out because you made an error in your calculations.

    For electricity to flow through R5, it has left the battery, passed through R4, gets to then through R5 then returns to the battery. R1,R2&R3 are not involved at all.

    14V applied to an 8 ohm branch means current of 14/8 amps.
    Through a 6 ohm resistor means 10.5 Volts drop across it.

    Another way is to consider R4 and R5 as a simple voltage divider circuit.

    In the text book 2 Ohm and 6 Ohm means V5 is 6/(2+6) * 14
     
  6. Sep 23, 2011 #5
    Ahh you're right. Thank you so much for your help.

    I was just starting to get convinced I needed to use all of the resistors...Much Obliged!
     
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