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Finding total resistance of circuits with cells in parallel

  1. Jul 23, 2004 #1
    I'm having trouble find the total resistance of circuits with cells in parallel. The answer in the book is 31 ohms, but my calcuation ends up as 30 ohms. What am I doing wrong? (Please show steps if not too much trouble) Thanks in advance!:D

    See this site for my work and a diagram of the circuit.


    The blurry thing says
    2 volts/cell
    Internal resistance: 1 ohm/cell
    And the resistors in the bottom left corner are 10, 5, and 15 ohms. Sorry for my crappy drawing, I did it on paint!
    Last edited: Jul 23, 2004
  2. jcsd
  3. Jul 24, 2004 #2


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    Assuming you did the rest right, you calculated the internal resistances wrong. It seems we have two cells in series, and that in parallel with another two cells in series. The two cells in series, each with 1 Ohm of resistance, will give 2 Ohms. Putting the two in parallel will give 1 Ohm for that part of the circuit. So you'd replace the 0.5 you've got with 1. If the rest is right, you should get 30.5, which they may have rounded to 31. Check your numbers. Also, I'm not entirely sure if this is the right way to deal with resistances of cells, but it's something to keep in mind, someone will probalby come along with more certainty soon.
  4. Jul 24, 2004 #3

    That's the most probable explanation I can think of right now, other than the book is wrong -_-

    According to the site

    "In general, when cells are connected in series, the total electromotive force (emf) is the sum of the emf's of each battery, and the internal resistance is equal to the sum of the separate internal resistances of the cells.

    When cells of equal emf and internal resistance are connected in parallel, the resultant emf is the same as that of one cell only and the internal resistance of the battery can be calculated from the formula for resistors in parallel."

    I think that series is when current flows through consecutively and parallel is when it flows through concurrently, and the two cells to me, seem like they're connected in parallel. According to the excerpt, I should just add the internal resistances as 1/( (1/1) + (1/1) )

    Maybe the cells are connected in series. *confused* Any help would be appreciated!
  5. Jul 24, 2004 #4


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    Your diagram shows 2 BATTERIES connected in parallel. Each BATTERY consists of 2 cells connected in series. So each battery has an internal resistance of 2 Ohms. The parallel pair of batteries has a total resistance of 1 ohm.
  6. Jul 24, 2004 #5
    Doh! *whacks self in the head* That would explain it, lol, thanks:D
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