Battery Problem - With Loop Rule (pic attached)

[aleksxxx]

Homework Statement

1. "lets set the internal resistance, $$r^_i$$, of the battery with emf $$E^_i$$ equal to $$bE^_i$$ where b is a constant assumed the same for all three batteries.

$$E^_1$$=5v
$$E^_2$$=10v
$$E^_3$$=15v
r1=b$$E^_1$$
r2=b$$E^_2$$
r3=b$$E^_3$$

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:
$$E^_1$$-I$$r^_1$$+$$E^_2$$-I$$r^_2$$+$$E^_3$$-I$$r^_3$$-I(200ohms)=0

After substituting in the EMFs and resistances i got the equation:
-2.6=-.137b$$E^_1$$-.137b$$E^_2$$-.137b$$E^_3$$

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=.6Im 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 I am going wrong here...i tried this and another way and i still get a deltaV higher than the combined voltages of the three batteries

 

[Mentz114]

I don't see how you get the first equation, or how you get from there to the second one.

 

[aleksxxx]

I don't see how you get the first equation, or how you get from there to the second one.

i used the loop rule and started at E1 and went clockwise to get the first equation.

Then i just plugged in the values of the batteries and the resistors to solve it.

I still cannont figure out what I am doing wrong becuase the voltage I am getting is way out of whack.

 

[Mentz114]

The first equation looks wrong. What about the emf across the 200 Ohm resistor ?

 

[aleksxxx]

The first equation looks wrong. What about the emf across the 200 Ohm resistor ?

There wouldn't be an EMF in that part of the equation for that, would there?

I thought the loop rule was, in the direction of the loop is you hit a battery - then + terminal, its positive that EMF, and when you go through a resistor if the loop and current are the same direction its -IR.

The loop doesn't go through another battery right before the 200 ohm resistor

 

[aleksxxx]

the other way i tried the problem was like this:

using the equation: E=IR
I took the EMF of the first battery, 5v, and set the equation as follows:

5V=(.137A)(r1) ==> where r1=bE1
5V=(.137A)(5V)(b)
b=7.3

Using that value in the second part of the problem still gives me a voltage drop of 82V across that 200ohm resistor, which can't be correct.