The first 2 images represent SERIES CIRCUITS. In a series circuit, the current is constant and is set by the total resistance of the circuit (the sum of the resistors). If you remove one resistor (or light bulb, as in the first image), the current stops. If the resistors were identical bulbs, having more bulbs would result in dimmer bulbs, since the battery voltage is distributed among them.

The last 2 images are PARALLEL CIRCUITS. Here, current has multiple paths to take, so the total resistance of the circuit is actually LESS than if the resistors were alone or in series with other resistors. Since the bulbs are connected equally to the battery, they experience the same as the battery voltage - they are, therefore, of equal brightness (and the same brightness they would have if there were only ONE bulb connected). Of course, bulbs in parallel draw more current and they cause a battery to die sooner.

What I've written above is primarily geared toward identical bulbs. In series, add up the resistances to get the total resistance. In parallel, it is more complicated. There is a formula one can use (1/Rp = 1/R1 + 1/R2 + ...), but we will only concern ourselves with the case of identical resistors in parallel. In that case, divide the value of the resistor by the number of resistors to get the total effective resistance. For example, two identical 50-ohm resistors in parallel is the same as one 25-ohm resistor. This seems strange, but it's a little like toll booths - when one toll booth is open, it can get crowded (the current is small). With multiple toll booths open, the resistance is effectively less, so the current can be greater.

The last 2 images are PARALLEL CIRCUITS. Here, current has multiple paths to take, so the total resistance of the circuit is actually LESS than if the resistors were alone or in series with other resistors. Since the bulbs are connected equally to the battery, they experience the same as the battery voltage - they are, therefore, of equal brightness (and the same brightness they would have if there were only ONE bulb connected). Of course, bulbs in parallel draw more current and they cause a battery to die sooner.

What I've written above is primarily geared toward identical bulbs. In series, add up the resistances to get the total resistance. In parallel, it is more complicated. There is a formula one can use (1/Rp = 1/R1 + 1/R2 + ...), but we will only concern ourselves with the case of identical resistors in parallel. In that case, divide the value of the resistor by the number of resistors to get the total effective resistance. For example, two identical 50-ohm resistors in parallel is the same as one 25-ohm resistor. This seems strange, but it's a little like toll booths - when one toll booth is open, it can get crowded (the current is small). With multiple toll booths open, the resistance is effectively less, so the current can be greater.