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Battery Problem - With Loop Rule (pic attached)

  • Thread starter aleksxxx
  • Start date
  • #1
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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]
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:
[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
 

Answers and Replies

  • #2
5,428
291
I don't see how you get the first equation, or how you get from there to the second one.
 
  • #3
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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 im doing wrong becuase the voltage im getting is way out of whack.
 
  • #4
5,428
291
The first equation looks wrong. What about the emf across the 200 Ohm resistor ?
 
  • #5
23
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The first equation looks wrong. What about the emf across the 200 Ohm resistor ?
There wouldnt 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 doesnt go through another battery right before the 200 ohm resistor
 
  • #6
23
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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 cant be correct.
 

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