As you can see here, to the left, the ammeters are demonstrating the fact that in a series circuit, the current is the same at every point. Why might that be? Well, let's look geometrically at the circuit. It is a closed loop, with only one path (around the circuit, from one end of the battery to the other). Since the electrons have only one path to travel, they can't get split up (as in parallel). So, the current remains the same everywhere because the constant flow is concentrated in only one path, therefore it is constant on that one path.
If we look at both pictures, specifically at the voltages, we will discover something else. The voltage of one light bulb (18.33V) is exactly one third of the voltage of the batteries (55V). It isn't a coincidence that there are three bulbs. What happens is this: the voltage supplied by the batteries is split evenly among the bulbs (since they have the same resistance). Unlike the constant flow of current, the voltage has three places to go across the circuit. Due to the current and resistances, the amount of voltage is determined with every resistor. This illustrates the conservation of energy (voltage).
The current is the same at every point in the circuit. An equivalent resistance can be calculated by adding the individual resistances of all of the resistors. The total voltage is equal to that of the battery, but can also be found by adding the voltages through each of the resistors.
The voltage, on the other hand, works differently. It is the same in every branch. The voltage on every branch is that of the battery, 19V in this scenario. This is due to the fact that the voltage has many paths to go across, which is the reason that it doesn't have to split. Although that reason makes the current split, the voltage works otherwise. Here, the voltage of every light bulb is the same, and equal to the battery. Every bulb has its own path for the voltage to go across, thus making every branch have the same voltage. The voltmeters in the pictures help to show this idea.
Of course, all of this depends on the resistance. Since all of the bulbs have the same resistance, and the batteries have none, the currents are split evenly. If the resistances changed, so would the currents although the voltages would stay the same since they don't depend on the resistance (and since if the resistance changes then so does the current, and when multiplied together, the result will always be the same).
The current is equal to the sum of the currents through each parallel branch. An equivalent resistance can be found by taking the inverse of the sums of the inverses of the individual resistors. The voltage is the same through every branch of the circuit.
Current works as a mixture of ideas from series and parallel here in a complex circuit. If we recall on some ideas, we know that from series, the current is the same every where, and that from parallel, it is spilt between branches. That is seen here. First we must treat the entire parallel segment as one part of a series. Then we see that the brighter bulb has 2.2A, or the total current as seen by next to the batteries. We also see that the entire parallel segment has 2.2A, but spilt between the branches, as explained earlier.
Thus, the current acts in a predictable manner. It is equal at every point, but under the condition that the parallel segment is considered one point.
The voltage is also another mixture. In series, it is split amongst the resistors. In parallel, it is evenly distributed to each branch. To observe this, one would have to look at all four pictures. The batteries supply a total of 33V. Now notice that the bulb in series gets a total of 22V, not the full 33V. Just like in series, no single bulb (if there are more than one) gets all the voltage. Then one can notice that the other two bulbs in parallel each have 11V. The voltage was split evenly across them, like predicted. But why in such a complicated pattern?
After the series bulb eats away at some of the voltage, there is only the remaining left for the parallel bulbs to consume. It is no coincidence that 33-22=11. After using some voltage, the series bulb left some remaining voltage left for each of the parallel bulbs. Both parallel bulbs have the same remaining voltage because they are in parallel (Read above). In the end, you will notice that the voltage consumed by the series bulb added to the voltages consumed by either one of the parallel bulbs results in the total voltage drop of the battery.
To find the equivalent resistance, add the equivalent resistance of the parallel branches and add that to the series resistor. The total current can be found by taking the voltage supplied by the battery and dividing that by the equivalent resistance. To find the voltage in the series resistor, multiply the total current by its individual resistance. Now to find the volatge in the individual branches of the parallel part, subtract the voltage eaten by the series resistor from the total voltage. To find the individual currents of the parallel resistors (unlike the series one, which has a current equal to the total one, as suggested by series), take the new found voltage and divide that by their individual resistances. If you will notice, the sum of the currents that are in the parallel branch equals the total current.