1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parallel Batteries.

  1. Mar 19, 2006 #1

    "Two automobile batteries are connected in parallel to power a wheelchair. If each of the batteries has an emf = 12.0 V and internal resistance r = .020 ohms, and the wheelchair motor has a resistance R = 1.00 Ohms, find the current provided to the motor. What would be the current delivered to the motor if the batteries were connected in series? What are the relative advantages of series and parallel connections?"

    Work thus far:

    I used 3 equations to solve for the current across the wheelchair.

    I(1) + I(2) = I(3) (Current In = Current Out)
    -I(1)r(1) + emf - I(3)R = 0 (Circuit loop 1)
    -I(2)r(2) + emf - I(3)R = 0 (Circuit loop 2)

    Solving for I(3) after 4-5 lines:

    I(3) = (2*emf*r) / (2r-2Rr)

    I(3) = -12.1 Amps.
    I wasnt really expecting a negative answer. I have the current over R from negative to positive, thus I'm not totally sure why its negative.

    I havnt done series yet because I wanted to confirm the work I did for // first.
  2. jcsd
  3. Mar 19, 2006 #2
    That tells me we are going to work with loops in this particular situations...

    Current is the same in series, Voltage is the same in parallel.

    Now, I cannot confirm your signs without a picture showing which way you decided to use for the current flow...
  4. Mar 19, 2006 #3
    Heres the picture (Roughly went thru paint)

    Attached Files:

  5. Mar 19, 2006 #4
  6. Mar 19, 2006 #5

    I figured out my problem, I just did some algebra wrong. the answer I came out with was:

    2Er / r^2 + 2Rr

    = 11.9A

    My problem now is that I have to compare to a series circuit set up.. but this leads me to the same problem I had on another thread. If they are in series I'm not totally sure how to find the amerage.

    What I did thus far was:

    Loop: -Ir + E - Ir +E - IR = 0

    I = 2E / 2r + R

    I = 23.1A

    I thought series would have less amps and more voltage.....
    Last edited: Mar 19, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Parallel Batteries.