Lamps with different values

Kirchhoff's Current Law definitely applies here. The sum of currents going into a junction equals the sum of currents leaving that junction. You can do the math and see it for yourself here. However, KCL won't do you much good trying to calculate current limiting resistor values. That is Ohm's Law, which states that the current passing through a conductor is directly proportional to the potential difference between its ends. So if you know the voltage entering a conductor, the voltage at the other side of the conductor (or at least what you want that voltage to be), you then know the potential voltage. Then since you also know how much current you want to pass, you can determine how much resistance you will need to make that happen. For example, you have an 18V 2.7W bulb. That means the bulb will need 150mA to operate properly. You have a supply voltage of 24V. So the potential difference of the resistor will be 6V, and you need it to pass 150mA. Resistance = Voltage / Current, so you'll need a 40 ohm resistor to pass 150mA with a 6V difference. That's all Ohm's Law. Kirchhoff's Law tells us that at the first node of the first resistor, there is 1117mA flowing in. Current is leaving the node in two directions. We already know that 150mA is going out one branch to the first lamp. So since current in = current out, then there must be 967mA going out the other branch. Which there is. Then at the next one, 967mA is going in, 667mA are going out one branch, so there must be 300mA out the other. Then the final one is easy, 300mA in, 300mA out. The two are closely related, but the resistor values are calculated using Ohm's Law.

@rbrtkurtz Thanks a lot for the analysis. That realy helped .