Battery Problem - With Loop Rule (pic attached)

In summary: The Attempt at a SolutionOk - For part one i set up the loop rule equation as follows:E^_1-Ir^_1+E^_2-Ir^_2+E^_3-I r^_3-I(200ohms)=0After substituting in the EMFs and resistances i got the equation:-2.6=-.137bE^_1-.137bE^_2-.137bE^_3From 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
  • #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=.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
 
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  • #2
I don't see how you get the first equation, or how you get from there to the second one.
 
  • #3
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 becuase the voltage I am getting is way out of whack.
 
  • #4
The first equation looks wrong. What about the emf across the 200 Ohm resistor ?
 
  • #5
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
 
  • #6
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.
 

1. What is the loop rule and how does it relate to battery problems?

The loop rule, also known as Kirchhoff's voltage law, states that the sum of the potential differences (or voltages) around a closed circuit loop must equal zero. This is important in battery problems because it helps us understand the distribution of voltage in a circuit and how it affects the flow of current.

2. How do I apply the loop rule to solve a battery problem?

To apply the loop rule, you need to identify all the components in the circuit and the direction of current flow. Then, starting at any point in the circuit, follow the direction of current flow and add up the potential differences across each component. The sum should equal zero according to the loop rule.

3. Can the loop rule be applied to any type of circuit?

Yes, the loop rule can be applied to any closed circuit, regardless of its complexity. It is a fundamental principle in circuit analysis that helps us understand the behavior of a circuit and solve problems related to voltage and current.

4. What are some common mistakes when applying the loop rule?

Some common mistakes include not considering the direction of current flow, forgetting to include all components in the loop, and not accounting for the signs of potential differences. It is important to carefully follow the steps of the loop rule to avoid these errors.

5. Are there any limitations to the loop rule?

The loop rule is based on the principle of conservation of energy and is a valid method for analyzing circuits. However, it does not account for the resistance of wires and other non-ideal components, which may affect the accuracy of the results. In some cases, other methods such as the node voltage method may be more suitable for solving circuit problems.

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