Not every circuit contains both parallel and series sections, but the ones in your exam probably will. Start with short sections that are clearly just series or parallel and merge these resistors together. Then, when the diagram is simplified, you will be able to combine the new resistors.
Key Concepts
Combining resistors in parallel is slightly more complicated than in series:
Resistors in parallel have the same potential difference across them. This is a result of Kirchhoff's second law, as the potential difference in any complete loop of the circuit must equal the EMF.
However, current splits at a parallel junction in inverse ratio to the resistance.
The resistance of a parallel combination is less than the individual resistors.
The equation is \({1\over R_T}={1\over R_1}+{1\over R_2}+...\)
NB: Think the algebra looks simple? Think again. You will need to combine the total right hand side before you flip.
The equation becomes easier with equal resistances - with two, you can simply half either resistance (and with three, take a third, etc).
Essentials
Potential difference across parallel resistors
When dealing with resistor networks calculate the resistance of the parallel bits first.
In which order should you approach the following circuit?
Combine series section R1 and R2
Combine series section R3 and R4
These will then become a parallel section - combine this
Combine your new resistor (containing R1, R2, R3 and R4) with R5 and R6
You then have one overall resistor in parallel with R7
What about light bulb filaments?
So far we have considered only simple resistors that produce thermal energy. If these resistors are replaced with identical light bulb filaments, that with the largest current will be brightest.
There are a couple of advantages to connecting several light bulbs in parallel, rather than in series:
A broken bulb can quickly be identified as it will be the only one to go out (and they can be switched on and off independently)
The bulbs will be brighter than the equivalent series combination as each has a direct connection to the EMF (NB: the circuit will use more electrical power, however)