Apparent contradiction in a simple circuit

[palaphys]

Homework Statement

Find the current across the cell E connected between A and B. given that E=2V and R=4ohm

Relevant Equations

Ohms Law, KCL, KVL

So I came across this simple looking circuit, and it seems to me that the uppermost branch and lowermost are parallel w.r.t AB.
applying the batteries in parallel formula, we get $$E_{net}=E $$ and $$R_{net}=R/2$$, simplifying the circuit as shown:

From here, it seems that the two cells (the original E in branch AB and the equivalent emf of the two parallel cells) are in series aiding connection, making the equivalent emf 2E, so current across AB comes out to be I=2E/(R/2)= 2amp. (as per the values given)

But, If we take a step back and apply basic nodal analysis, taking the potential at A as 0, we get a circuit like this, which seems to tell us that no current flows through the branch AB, and a current of 2E/R is flowing through the lower and upper branches in an anticlockwise sense.

What's gone wrong, where, and why?

 

[berkeman]

But, If we take a step back and apply basic nodal analysis, taking the potential at A as 0, we get a circuit like this, which seems to tell us that no current flows through the branch AB, and a current of 2E/R is flowing through the lower and upper branches in an anticlockwise sense.

Since the voltage across each resistor is 4V, what is the current in each of the outer branches? And those currents add to make the 2A that flows back through the middle branch left-to-right.

 

[kuruman]

But, If we take a step back and apply basic nodal analysis, . . .

Let's take a few more steps back and apply the two basic conventions about ideal DC circuits:

1. Batteries maintain a constant potential difference across their terminals;
2. Straight line segments are equipotentials.

Let's ground the negative terminal of the top battery to declare the potential at that point zero (see figure on the right.)

Starting at this zero and going counterclockwise, we see that all straight line segments connecting point A to each of the three batteries will be at the same electrostatic potential of 2 V.

Now he battery in the middle branch raises the potential of point B by another 2 V so that all the straight line segments connected to B are at 4 V. Now if you look at the potential difference across each 4 Ω resistor, you will see that the current through each must be 1 A to the left, from high to low potential.

Finally, current conservation at nodes A and B demands that the current in the middle branch be 2 A to the right as already noted by @berkeman.

The last circuit you drew is consistent with mine. This is where you misled yourself

and a current of 2E/R is flowing through the lower and upper branches in an anticlockwise sense.

Both currents are left to right, from high to low potential. Left to right is counterclockwise in the upper branch but clockwise in the lower branch.

 

[palaphys]

The last circuit you drew is consistent with mine. This is where you misled yourself

Both currents are left to right, from high to low potential. Left to right is counterclockwise in the upper branch but clockwise in the lower branch.

so there was no contradiction, and I wasn't wrong with my diagrams, just a silly conclusion which I overlooked.

Thanks, I got it