ObamaOrHillary2008 wrote: (snip)
Each transistor is provided a potentrial base drive through a series combination of the 1k collector resistors and the
4.7k base resistors (neglecting the LED for this description).The base to emitter junctions start to pass current when that junction has as little as .6 volt across it. So the remaining 2.4 volts is divided between the 1k and 4.7 k resistors, with the 1k dropping about 2.4*1k/(1k+4.7k)=0.42 volts and the 4.7k dropping about 2.4*4.7k/(1k+4.7k)=1.97 volts. If neither transistor collector was collected to anything, this is all that would happen, with the junction of the two resistors having a voltage of about the .6 volt base drop plus the 1.97 volts across the 4.7k resistor for a total of 2.57 volts. Disconnect the collectors and the LED and see if this is what happens.
When the circuit is first powered on, both these resistor pairs start to turn on their respective transistor (which I have labeled Q1 and Q2 on your drawing to make references easier) by providing some base current.
But as each transistor begins to conduct collector current, that additional current through the 1k collector resistor increases the voltage drop across them, lowering the voltage feeding base current to the opposite transistor. This tug of war ends with one transistor being on (low collector voltage, with most of the 3 volt supply dropped across its
1k collector resistor and about 3 mA passing through the collector to ground) and one being off, with only the base current for the other transistor passing through its collector resistor and the collector voltage at about 2.57 volts.Lets say that the on transistor is Q1. So Q1 will see a base voltage of about .6 volts while it will short out the source of base current for Q2, so Q2's base voltage will be some small fraction of a volt that is seen at the collector of Q1. There will be no voltage drop across Q2's base resistor because the base voltage is too low for that transistor to draw any base current.
At that point, shorting the small fraction of a volt at the base of Q2 to ground with switch S does almost nothing, since Q2 remains off, and the tiny current to ground through the 4.7k and switch only helps to lower the already low voltage at the collector of Q1 by a millivolt or so.
But if you close switch R, shorting to ground the significant base current that is being delivered to Q1 by its base resistor from the 2.57 volts at the collector of Q2, you immediately turn Q1 off. This allows the voltage at its collector to rise to 2.57 volts, which turns on Q2. Once Q2 is on, it detours the base drive for Q1 to ground, so the R switch no longer does much.