- #1

- 23

- 0

## Homework Statement

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.

[tex]E^_1[/tex]=5v

[tex]E^_2[/tex]=10v

[tex]E^_3[/tex]=15v

r1=b[tex]E^_1[/tex]

r2=b[tex]E^_2[/tex]

r3=b[tex]E^_3[/tex]

http://i17.photobucket.com/albums/b51/chs2004/physics1.jpg

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.

http://i17.photobucket.com/albums/b51/chs2004/physics2.jpg

## Homework Equations

Kirchoffs's (sp) Loop Rule

## The Attempt at a Solution

Ok - For part one i set up the loop rule equation as follows:

[tex]E^_1[/tex]-I[tex]r^_1[/tex]+[tex]E^_2[/tex]-I[tex]r^_2[/tex]+[tex]E^_3[/tex]-I[tex] r^_3[/tex]-I(200ohms)=0

After substituting in the EMFs and resistances i got the equation:

-2.6=-.137b[tex]E^_1[/tex]-.137b[tex]E^_2[/tex]-.137b[tex]E^_3[/tex]

From there i factored out the .137 and brought that to the other side and then factored out the 'b' and got:

19=b(5+10+30)

b=.6

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