(03-03-2017, 03:46 AM)Nowhere-Man Wrote: 4.3 would be 9 volts minus the 4.7 dropped by the 4700 load. There's another observation. Ok I got it. Let's assume we know the "tables" gave us 4.3 collector volts and supply is 9 volts.
NO! forget the "tables" for now, the answer was not in them.
You have said the correct answer.
4.7v is the voltage dropped across the 4.7k resistor because it has 1mA flowing through it. V=IR (I is the current)
So the voltage at the collector, (with reference to 0 volts, which is maybe what is confusing you), is 9v - 4.7v = 4.3 volts
The answer Mark was looking for was in your first sentence of your last post.
So to recap the complete answer to Marks original question is b = 1.6v, e = 1v, and finally c = 4.3v
Please, NWM can you go over this again to check how each of the 3 voltages were established, and ask any relevant questions you want if you have any problems.
Mike
