This is because the drop in voltage across L1 is, just the difference between the saturation voltage of one transistor and the base-emitter voltage of the other transistor. Roughly speaking that is going to be about half a volt. With a 220 Ω resistor that is going to amount to a reduction of about 2.3 mA (you are seeing a reduction of 3.3 mA, which is well within that ballpark).@dl324 - your circuit doesn't seem to work, in Every circuit, with the switch closed L1 uses 13.4mA and L2 0mA.
But with the switch open, L2 gets 13.5mA but L1 still gets 10.1mA.
As you draw more total current you get more voltage drop across the shared transistors and, therefore, less current.There is one thing is puzzling me. All the transistors for the "switches" are in parallel. So why does turning them on one at a time decrease the current draw across the LED's that are lit?
http://everycircuit.com/circuit/6146264910528512

