Supply voltage of 15V goes through R1 of 10K Ohms that leaves only a small amount of available voltage enough to push the small current through b-e, in order to make the connection between c-e. This is the part where a transistor operates as a switch.
Because as we vary the supply voltage or the resistor before base, voltage drop @ base will slightly change, plus we can't really determine the value of the resistance between a likeliness of diode, then a voltage drop will come in handy:
As 15V goes through R1 and goes through b-e, the total voltage drops between these 2 will add up to near 15V. We know the VD of the b-e is 0.797V, then we can get the Voltage Drop across the resistor: 15 - 0.797 = 14.203V. Hence the Current through base: I= Vr1/R1= 14.203/10000= 1.4mA approx~.
Knowing the base current is one of components which help determine how big the amplified current will be. Since VD across c-e is impractical to determine using the same method, we will have to advance deeper into the operation of the transistor.
However, on the practical field, we can always measure using Multimeter- connected across collector and emitter, and a reading of 46.3mV obtained. When a small current goes through base, a low voltage drops between c-e indicates that very small effort it takes to conduct as this act almost as effective as a conductor, hence this really low resistance allows a bigger, amplified current to go through.
A- Saturated Region. This is when the transistor works at its best. Only a small increase in Ib(which is the Base current) can result in a steeply effective Ic( Amplified Current) switched on and amplified. (Take a point on the load line in A region aims straight down the Ib line (which is the base current)). Also, we can see that only a VERY small amount of Vce (voltage drop) across what supposes to be nearly as good as a conductor, then we have ourselves a transistor working @ in other word:" turned on at maximum level.
B- I mention "near Cut-off" because it is supposed to be below the forward base current level. This "cut-off" means similar to "the whole city of New York's electricity line has been cut off for several hours for maintenance due to excessive usage". It's a "cut-off", where no current flows through the base as the transistor is simply switched off.
Active Region: In between Saturation and Cut-off. This is the range of Ib and resultant Vce where the transistor is not operating at its best, usually when varying between wide range of resistor configurations which effectively alter the current outcome. As we can see a small rise in base current leads to a very significant rise in voltage drop between what we need as a conductor(Vce).
Power dissipation of a transistor:
Our transistor is physically a component on a circuit board but NOT necessarily in term of "loads" or "consumers" etc... In fact, it's a JUNCTION which when operates will eventually have 2 loaded consumers: They are: Base-emitter and Collector-emitter. Each component has its own Voltage drop( Vbe; Vce) and Operating Current( Ib; Ic), therefore we actually have 2 components which have power dissipation. P= Pbe + Pce= Vbe x Ib + Vce x Ic.
Beta: These are the amplified current gain from the base current, calculated by: B= Ic/Ib.
Where in saturated region the transistor will experience maximum current gain.
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