Circuit battery current Problem

[clope023]

Homework Statement

http://img87.imageshack.us/img87/6702/yffigure2534pk3.jpg

The circuit in the above figure contains two batterires, each with an emf and an internal resistance, and two resistors. Find a) the current in the circuit (magniturde and direction) b) the terminal voltage Vab of the 16V battery c) the potential difference Aac of paoint a with respect to point c

Homework Equations

I = V/R

V = \epsilon - Ir

Vac = Vab-Vbc

The Attempt at a Solution

I1 = V/R = 16V/1.6ohms = 10A (first battery)

I2 = V/R = 8V/1.4ohms = 5.7A (2nd battery)

Vbc = \epsilon - IR = (16+8V) - (10A + 5.7A)(9ohms) = -117.3V

Vab = \epsilon - IR = (16V) - (10A)(1.6ohms) = 0V

Vac = Vab-Vbc = 0 - (-117.3V) = 117.3V

I'm not quite sure if I did the problem correctly, if anyone could check my method I would greatly appreciate it.

 

[Mindscrape]

I don't know if I follow your method, but here's the method I would use:

Sum of voltages around the circuit loop is zero, as Kirchoff's Voltage Law dictates. So starting at node 'a'...

1.6I-16V+9V+...(you finish)=0

 

[clope023]

I don't know if I follow your method, but here's the method I would use:

Sum of voltages around the circuit loop is zero, as Kirchoff's Voltage Law dictates. So starting at node 'a'...

1.6I-16V+9V+...(you finish)=0

um, I understance where you get the intitial resistance and voltage but where does 9V come into play?

would it be:

I(1.6ohms) + 16V - I(9ohms) - 8V - I(1.4ohms) + I(5ohms) = 0 and then I could solve for I?

 

[clope023]

using the setup I made using the Loop Rule I got I = 2A through the circuit, does this seem correct?

 

[clope023]

okay well using my attemtped model, I did the terminal voltage for the 16V battery

Vab = E - Ir = 16 - (2*1.6) = 12.8V

Vac was done taking what looked like the easier path first going through the 5ohm resistor and then through the 8V emf and the 1.4ohm resistor with the 2A current and did:

(-2A*5ohm) + 8V + (2A*1.4ohm) = Vac = .8V

anyone?

 

[Dick]

Why are you making some of the voltage drops across resistors positive and some negative? You should be assuming the current is flowing in the same direction through the whole circuit.

 

[clope023]

Why are you making some of the voltage drops across resistors positive and some negative? You should be assuming the current is flowing in the same direction through the whole circuit.

I assumed I could pick direction and going below the a node I would be going in the negative direction going through the 5ohm resistor.

 

[Dick]

Here's what I would do. And I'm not very good at this because I confuse plus and minus a lot but let's assume current is flowing counterclockwise and put a voltage drop across each resistor. Start at a. -5*I-1.4*I-8-9*I+16-1.6*I=0. Am I doing ok, or did I get a sign wrong?

 

[Mindscrape]

Ah, yes, I meant to say 9I. So going from point 'a' all the way around the loop, and assuming a clockwise current, gives.

1.6I-16V+9I+8V+1.4I+5I=0 (the same as Dick's if you multiply by through by -1)
18I=8
I=4/9 and clockwise

Does it make sense to you?