How to determine direction of current mentally in complex electrical networks? I wanted to ask, as stated above, how would one work out, by considering such thing as potential differences provided by the battery, the path of the current in a complex network such as given below:
 
Would current start flowing from the positive terminal of $E_2$, will it then divide into two parts and pass through $R_1$ and $R_5$ and what would be the general sense of the current (counterclockwise/clockwise)?
How can all these questions be answered intuitively rather than by application of the Kirchoff's Rules?
 A: It depends a lot on the circuit. Some circuits are insanely complex to such an extent that it is next to imposible to do the calculations mentally. In fact, most of us use simulators to solve complex circuits.
Anyway, the best tool you've to work on simple resistor & multi-battery circuits mentally is superposition principle. 

The Superposition Principle
According to the principle, you can treat a circuit containing many sources to be made up of several different circuits with single source each. You can solve the circuit containing multiple sources by considering each source to be acting alone and add the effects of all the sources to get the final answer. The following example will illustrate the idea.
Lets say you had to solve (find the currents in each branch) the following circuit,

According to the superposition principle, the circuit can be thought to be made up of two separate circuits each with one of the source.
We are basically selecting a particular source and remove rest of the sources to get a sub-circuit. We keep doing until we have a sub-circuit for each source. In our case, there are two sources and hence there shall be two sub-circuits.

The two battery circuits is now a problem consiting of two separate single battery circuit. Solve each of them separately and you get the currents in each case to be the following,
For the first circuit (at the left),
$I_{R1} = 1.5A\space$ downwards, +ve terminal of the battery to the -ve terminal
$I_{R2} = 3.0A\space$ downwards, +ve terminal of the battery to the -ve terminal
For the second circuit (at the right),
$I_{R1} = 1.5A\space$ upwards, +ve terminal of the battery to the -ve terminal
$I_{R2} = 0.0A\space$ shorted
All that is left is to add the corresponding currents from the two sub-circuits to get the current through the resistors of the combined circuit. If the current is in opposite directions, then they will cancel as much as possible and the net current will be in the direction of the sub-circuit current which was higher.
So the final answer will be,
$I_{R1} = 0.0A\space$ This was expected. Wasn't obvious? Think why.
$I_{R2} = 3.0A\space$ downwards
This idea can be extended to circuits containing any number of sources.
