Parallel Batteries: Power a Wheelchair Motor

[moo5003]

Problem:

"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.

 

[Da-Force]

"Two automobile batteries...

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...

 

[moo5003]

Heres the picture (Roughly went thru paint)

 

[Da-Force]

Use..

http://pics.xs.to/

to upload picture.

 

[moo5003]

http://xs.to/xs.php?f=CircuitSketch010101.JPG&h=xs73&d=06121

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...