Current crossing the bridge

The problem can be solved very easily by using the method of superposition. The goal is to determine how each of the current or voltage sources affect the current we're looking for. It works this way:

• Pick a current/voltage source
• "Turn off" the rest of the sources in the circuit(replace ideal E-sources with short circuits and remove I-sources)
• Determine the current through desired branch when only the source you picked is turned on
• Repeat first three steps until you have picked all of the sources and calculated the currents they yield through the desired branch.
• Algebraically sum up all the yielded currents, by putting + in front of a current whose direction of flow is the same as the assumed direction of flow of the final current(given in the picture).
• After summation, you'll get the current you're looking for.

There is one current source and two voltage sources, so we have three cases:

Given the above rules, we have that: $I_X = I_{X1} + I_{X2} + I_{X3}$ Take a look at the first case. We can rearrange it this way: So, we have three identical resistors connected in parallel to the current source, which means that the current flowing from the source will be split into three equal parts. Thus: $I_{X1} = \dfrac{I}{3}$

Now, take a look at the second and the third case. Notice that you get the third case if you rotate the second case by 180°. It means that second and third case yield currents of equal intensities that flow in opposite directions - so they cancel each other out: $I_{X2} = -I_{X3}$

Having all this in mind, we finally have that : $I_X = I_{X1} + I_{X2} + I_{X3} = I_{X1} = \dfrac{I}{3} = 4A$ .

Step 1 - Create a "super node" as shown by the green ellipse. The sum of the currents out of this must be zero.
Step 2 - Symbolically label the node voltages as shown
Step 3 - Write an expression for the total current flowing out of the "super node"

$\frac{V-E}{R} - I + \frac{V}{R} + \frac{V+E}{R} = 0 \\ V = \frac{IR}{3} \\ I_X = \frac{V}{R} = \frac{IR}{3R} = \frac{I}{3} = \boxed{4}$

Interestingly, we don't actually need the numerical value of $E$ or $R$ .