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Specialty Chemistry Forums => Other Sciences Question Forum => Topic started by: aleksxxx on March 11, 2007, 01:05:15 PM

Title: Battery Problem - With Loop Rule (pic attached)
Post by: aleksxxx on March 11, 2007, 01:05:15 PM
Here is the problem

"lets set the internal resistance, ri, of the battery with emf Ei. equal to bEi where b is a constant assumed the same for all three batteries.

E1=5v
E2=10v
E3=15v
r1=bE1
r2=bE2
r3=bE3


(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi17.photobucket.com%2Falbums%2Fb51%2Fchs2004%2Fphysics1.jpg&hash=41713eeca0d7305633b474aaac9d6af599f01faf)
http://i17.photobucket.com/albums/b51/chs2004/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.

(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi17.photobucket.com%2Falbums%2Fb51%2Fchs2004%2Fphysics2.jpg&hash=9e2a37b03c52a0c6786ed0ca63996753e1ddc2e0)
http://i17.photobucket.com/albums/b51/chs2004/physics2.jpg (http://i17.photobucket.com/albums/b51/chs2004/physics2.jpg)

2. Relevant equations
Kirchoffs's (sp) Loop Rule


3. The attempt at a solution

Ok - For part one i set up the loop rule equation as follows:
E1-(Ir1)+E2-(Ir2)+E3-(Ir3)-I(200ohms)=0

After substituting in the EMFs and resistances i got the equation:
-2.6= -.137bE1 - .137bE2 - .137bE3

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



----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.
Title: Re: Battery Problem - With Loop Rule (pic attached)
Post by: Borek on March 11, 2007, 05:52:43 PM
1. Where have you lost I when solving for b? There were two unknows (b & I) yet somehow you have calculated b value from the equation. Or was I given, you just forgot tp mention it?

2. Please write all equations describing the circuit. There are 4 unknowns - total current and three 'subcurrents" in three branches. Thus you need four equations. Or there are 5 unknowns (4 currents plus b) but I think you have to treat bas a parameter and compare total currents in both cases as a function of b.