1. "lets set the internal resistance, [tex]r^_i[/tex], of the battery with emf [tex]E^_i[/tex] equal to [tex]bE^_i[/tex] where b is a constant assumed the same for all three batteries.
2.To decrease internal resistance we can connect the batteries in parallel, although this advantage could be offest by a decrease in net voltage. To test this hypothesis find the current flowing through the 200 ohm resistor, assume 'b' is the same from part one.
Kirchoffs's (sp) Loop Rule
The Attempt at a Solution
Ok - For part one i set up the loop rule equation as follows:
After substituting in the EMFs and resistances i got the equation:
From there i factored out the .137 and brought that to the other side and then factored out the 'b' and got:
Im not sure i did that first part right, becuase for the second part when i figure out all the currents, I1, I2, I3, they all come out to the same thing becuase the equation, I=E/R, and R=bE, the EMFs just cancel giving me 1/b for each current, 1.7A=I1=I2=I3.
Then using I=5.1A solving for the deltaV across the 200ohm resistor i get:
DeltaV=(5.1A)(200ohms)=1020V (WAYYY too high)
Im not sure where im going wrong here...i tried this and another way and i still get a deltaV higher than the combined voltages of the three batteries