Battery Problem - With Loop Rule (pic attached)

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Homework Help Overview

The discussion revolves around a problem involving batteries with internal resistance and the application of Kirchhoff's Loop Rule. The original poster presents a scenario with three batteries of different EMFs and explores the implications of connecting them in parallel to decrease internal resistance while questioning the resulting current through a resistor.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the loop rule equation and question the validity of the original poster's equation. There are attempts to clarify how the EMF across the 200-ohm resistor should be accounted for in the equations. Some participants express confusion about the calculations leading to unexpected voltage results.

Discussion Status

The discussion is ongoing, with participants actively questioning the setup and calculations presented. There is no explicit consensus on the correctness of the initial equation or the subsequent calculations, and multiple interpretations of the loop rule are being explored.

Contextual Notes

Participants are working under the constraints of the problem as posed, including the assumption of a constant internal resistance factor 'b' for all batteries. There is a noted concern regarding the high voltage drop calculated across the resistor, which raises questions about the validity of the approaches taken.

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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=bE^_1
r2=bE^_2
r3=bE^_3
physics1.jpg

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.
physics2.jpg

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-Ir^_1+E^_2-Ir^_2+E^_3-Ir^_3-I(200ohms)=0

After substituting in the EMFs and resistances i got the equation:
-2.6=-.137bE^_1-.137bE^_2-.137bE^_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, because for the second part when i figure out all the currents, I1, I2, I3, they all come out to the same thing because 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
 
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I don't see how you get the first equation, or how you get from there to the second one.
 
Mentz114 said:
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 because the voltage I am getting is way out of whack.
 
The first equation looks wrong. What about the emf across the 200 Ohm resistor ?
 
Mentz114 said:
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
 
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.
 

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