Find terminal voltage, currents of multiple batteries.

[Blu3eyes]

Homework Statement

Find terminal voltage, currents of each battery
[PLAIN]http://img26.imageshack.us/img26/8667/questionl.jpg

Homework Equations

$$\epsilon=I(R + ri)$$
V_terminal= $$\epsilon$$ $$\pm$$ I*ri

The Attempt at a Solution

Find current:
I=$$\frac{\epsilon1+\epsilon2+\epsilon3}{R+r1+r2+r3}$$

Fine Voltage
V_terminal1= $$\epsilon1$$ $$\pm$$ I*r1

This is all I can think of and unfortunately, it does not seem right.

 

[gneill]

Careful, not all the batteries are in the same direction going around the loop.

 

[Blu3eyes]

Careful, not all the batteries are in the same direction going around the loop.

Yep, that is why I put the $$\pm$$ when finding the terminal Voltage.
Could you give me some suggestions??
I am thinking of using Kirchhoff's Laws but do not know how to start since all batteries are connected in series.

 

[gneill]

I would suggest finding the current first, ignoring whether the resistors represent internal or external resistances. Just assume a current direction and "walk" around the loop to write the loop voltage equation.

PS: You were *almost* right with your previous attempt, but if you look carefully at ε2 in the circuit, it faces in the opposite direction to the others.

 

[Blu3eyes]

ok, I would assume the counter clockwise current starting from emf 1. Current enters resistors with negative (-) end leaves with positive (+).
Loop :
emf1 - Ir₁ + emf3 - Ir₃ -Ir₂ - emf2=0
6 - 6I + 8 - 8I -16I -12 =0
30I=2
I=0.066667 A

Then find terminal voltage:
Vter = emf (Plus or minus?) Iri ??

 

[gneill]

I don't see your load resistor in there.

 

[Blu3eyes]

I don't see your load resistor in there.

Let's try it again:
emf1 - Ir1 + emf3 - Ir3 -Ir2 - emf2 - I*Rload=0
6 - 6I + 8 - 8I -16I -12 -370I=0
400I=2
I=5x10^-3A
V_t1=6V-5x10^-3A$$\times6\Omega$$=5.97V

V_t3=8V-5x10^-3A$$\times8\Omega$$=7.96V

V_t2=12V-5x10^-3A$$\times16\Omega$$=11.92V

Would you confirm, please??
Thanks!

 

[gneill]

You're 99% there. You've been tripped up again by the backward connection of emf2. That battery is actually sinking current rather than supplying it (conventional current of positive charges is flowing into its + terminal rather than out of it). So you need to add the potential across the internal resistance in this case.

 

[Blu3eyes]

You're 99% there. You've been tripped up again by the backward connection of emf2. That battery is actually sinking current rather than supplying it (conventional current of positive charges is flowing into its + terminal rather than out of it). So you need to add the potential across the internal resistance in this case.

I've got it. V_t2 should be 12.08.
Is there anyway I can check my work just so I know that I did not mess up the (+) or (-)??

 

[gneill]

I've got it. V_t2 should be 12.08.
Is there anyway I can check my work just so I know that I did not mess up the (+) or (-)??

If you add up all the battery terminal voltages (paying attention to polarity) and divide by the load resistance, you should get the current value that you determined earlier.