Determine which circuits are parallel and which are in series 
I need to find the Equivalent Resistance across A and B. The problem is, I don't know which one's are in series and which ones are in parallel. How do I identify which ones are parallel or series?
 A: 
How do I identify which ones are parallel or series?

If all of the current leaving one resistor enters another resistor, the two resistors are in series.
The resistances of series connected resistors can be added together to find the equivalent resistance of a single resistor, e.g.,
$$R_{eq} = R_1 + R_2 $$
If all of the voltage across one resistor is across another resistor, the two resistors are in parallel.
The conductances of parallel connected resistors can be added together to find the equivalent conductance of a single resistor, e.g.,
$$G_{eq} = \frac{1}{R_{eq}} = G_1 + G_2 = \frac{1}{R_{1}} + \frac{1}{R_{2}}$$
or
$$R_{eq} = \frac{1}{\frac{1}{R_{1}} + \frac{1}{R_{2}}} $$
A: *

*Two resistors on the same path are in series. There must be nothing else in between.

*Two resistors that follow parallel paths are in parallel. If the wire splits in to paths and gatheres into one again alter, this would be parallel paths.


In your circuit, you must piece by piece collect resistors into one equivalent resistor.


*

*Start with the $1.00\;\Omega$, $2.00\;\Omega$ and $3.00\;\Omega$ ones most to the right. They are in series, so they are added:
$$R_{eq}=1.00\;\Omega+ 2.00\;\Omega +3.00\;\Omega=6\;\Omega$$
If you drew the whole circuit again, you could replace these three with one resistor with this value.

*Now, this new equivalent resistor is in parallel with the vertical $3.00\;\Omega$ resistor. So they are added reversely:
$$\frac1{R_{eq}}=\frac1{3.00\;\Omega}+\frac1{6.00\;\Omega}=\cdots$$
These two can now be replaced with one new resistor of this value.

*Next step would fx be to see that this new equivalent resistor is in series with the $6.00\;\Omega$ one on top. So add them...
Continue in this way, until all are shortened down and there is only one equivalent resistor left. That will the be the resulting equivalent resistance between A and B.
A: You want to work this out in stages. Start at the right hand side where there are three resistors clearly in series. "In series" means that the current that flows through one resistor is equal to the current flowing through the other resistor. The simplest way to identify this is to look for nodes (points where components are connected) with just two wires: those nodes connect components in series.
Figure out their equivalent resistance (1+2+3).
Now that 6 ohm resistance is parallel to the remaining 3 ohm resistor - replace these two with a single resistor.
And so it continues. every time you find a combination that either looks like a number of resistors in series, or two resistors in parallel, you replace them with their equivalent.
Perhaps this picture makes it clearer:

The red dots correspond to points (nodes) where one wire goes in and one wire comes out. This indicates "series" connection. The two green dots inside the gray box correspond to nodes where three wires meet: current from the 6 ohm resistor can go either into the 3 ohm or the 1 ohm in the top green circle, and currents from the two 3 ohm resistors are joined together before going to the other green circle where they combine with the 4 ohm resistor.
If you can figure out how to combine everything in the gray box into a single resistor, then you can draw that single resistor instead of the four that are inside that box (their equivalent resistance is 2 ohms - that should help you check your work) and that resistor is then in series with the 6 ohm... So the two could be replaced with an 8 ohm...
A: Since this is a homework type question, I can only offer you a tip and not solve it for you.
Tip - Look at this series combination for example

Notice that there is some current $I$ flowing through each of these three, and the value of potential at points 1, 2, 3 and 4 (across $R_1$, $R_2$ and $R_3$) aren't the same. 
On the other hand, in a parallel connection: 

That's probably enough hint. You can now build on this :)
A: The answers are really good, but I have (as a former long-time student who has not forgotten the practicality of things) a more practical approach without considering currents or voltages, just the topology of the circuit itself:
When trying to reduce a circuit, the algorithm is as follows: 


*

*Any number (> 1) of resistors on the same wire. That is, no branches from the middle, just on the same wire. Then, these are in series. They are reduced to the sum of their resistance values. Repeat until there are no more multiple resistors on the same wire.

*Any number of resistors whose ends are connected to the same two points are in parallel now. Since you reduced all the series ones in the first step, now there can only be sets of single resistors with both ends tied to the same two points. For each such set, remove all resistors, and replace them with a single resistor between the same two points, whose resistance value is the inverse of the sum of the inverses of all the resistors in the set. 

*Go back to the first step, since parallel reductions may have created more resistors in series. If nothing has changed since the last time you reached this step, the circuit can not further be parallel-series reduced, and you are done.
Note that some circuits can not be parallel-series reduced at all. In that case, an all-resistor circuit can still be reduced, but that would require some circuit analysis; you can not get away with simple consideration of the topology and resistor values.
A: Just look at the nodes because they have a lot to say.
Two resistors are Series with each other with the following condition:
1- Have only one common node.
2- The common node should be connected to only two resistors.
Two resistors are Parallel to each other with the following condition:
1- Have only two nodes in common.
