net emf and resistance

Question

Find the net emf and internal resistance across AB of three batteries as shown in figure

 

figure is on http://i.imgur.com/Evu27EH.jpg

Answer 1

At first glance, I do not think there is enough information to solve here for either the net EMF or net internal resistance. We could work with the resistors alone, and find the net resistance, but having the batteries in the branches with them alters the current and thus the result.

The trouble, as presented, is that it looks like we have three unknowns -- the currents through the batteries -- and only two independent equations to relate them. If I call the currents i₁ (through the 3 V battery on the left), i₂ (through the 8 V battery above), and i₃ (through the 10 V battery below), then I can set up two relationships between the three currents.

1) From Kirchhoff's junction rule: i₁ = i₂+i₃
NOTE: Here, I guessed all currents flowing to the right in the diagram, which is probably not correct. But that would work itself out in the solution; I would get a negative sign in front of any current where I guessed it was flowing in the wrong direction)

2)  From the fact that the total potential difference across each branch in parallel must be the same

-i₂(2 Ω) + 8 V = -i₃(2 Ω) - 10 V  [NOTE:  this expression can also be obtained from Kirchhoff's loop rule around the loop containing the parallel branches]

Here, I am proceeding from left to right across each parallel branch, and assuming all currents flow to the right, as before. I use the rules that, if I traverse a resistor in the direction of current, I subtract its voltage (=IR). I also use the rule that, if I step from the minus (low voltage) to the plus (high voltage) terminal of the battery, as in the upper parallel branch, I add its voltage; and if I step from the plus to the minus terminal, as in the lower parallel branch, I subtract its voltage.

 

But, as far as I can tell, that is all that is known without more information.  Unfortunately, that leaves two equations with three unknowns, and I would need those unknowns to solve for what is asked.  If there is more information you can give, or if someone else has more insight with the given information, I would be happy to hear it and work further.

 

Assuming the net drop between A and B is 2 V, we can now set up a third relationship between the currents (one of two possibilities).  Again, all currents are assumed to the right in the diagram:

 

-- -i_i(1 Ω) + 3 V - i₂(2 Ω) + 8 V = -2 V

 

Now we have three equations relating the three unknowns.  With these, we can solve for the currents in each section.  With some algebra, I obtained:

 

i₁ = 4.2 A

i₂ = 8.8 A

i₃ = -4.6 A  (meaning i₃ actually flows from right to left in the diagram, opposite my initial guess... as I expected at least one of the currents to do)

 

As to what this means for "net resistance," I find it had to say.  In this context, the overriding principle is that the terminal voltages in each parallel branch have to be identical, which is why i₃ has to flow to the left.

 

As I mentioned, if the batteries were capacitors, we could collapse these into one equivalent resistor and one equivalent capacitor.  But batteries change the game, because they hold their own internal voltage due to their electrochemical centers, and affect current directly.  So I do not think it is fair to collapse all the resistors into one in this case.

 

I would need to ask the teacher a question about the purpose of the problem at this point.  The work up to now is a complete description of the circuit with the information given, but I would not be sure how to define "net resistance" in this context.  Sorry!