Electronic – How is the current through resistor same as collector current in this circuit

Dave is correct... I'll try to clarify some more.

In an NPN:

• The base-emitter voltage and the doping of the base determine the rate of emitter electron current injection into the base, which is swept into the collector due to the potential drop from base to collector and the narrowness of the base region.
• The base-emitter voltage and the doping level of the emitter also determine the rate of base hole injection into the emitter, which does reduce the collector current.
• The ratio of the dopant densities sets the current ratio between the base and collector (Beta).

BJTs are designed with light doping in the base and a very narrow base width to maximize the diffusion of the emitter current to the collector. As a result, the base current needed to develop the Vbe for a given rate of emitter current injection is very small compared to the emitter and collector current, and so BJTs have high current gain.

Here's an online reference that goes into some detail: Modern Semiconductor Devices for Integrated Circuits, Ch. 8

You cannot make both assumptions about \$V_{ce} = 0.2V\$ and \$\beta = 100\$ at the same time; they are for two different modes of operation for the transistor (there's also a third mode where the transistor is in cutoff, and a fourth less common one in reverse active).

You start out assuming one of the possible states. For example, suppose I assume the transistor is saturated. Then \$V_{ce} = 0.2V\$ and \$V_{be} = 0.7V\$. You can then solve for base and collector currents using these assumptions for the BJT and only these assumptions.

The last step is to check your assumptions. For example, to make sure the transistor isn't actually in the linear active region must check that \$\beta I_{b} \gg I_{c}\$. So let's say in your case you get \$I_{c} = 9.8 mA\$ and \$I_{b} = 9.8 \mu A\$. Well clearly our assumption about saturation was bad because \$\beta I_{b} = 980 \mu A\$, which is less than our calculated \$I_{c}\$. We must then start over with new assumptions and re-solve the problem.

Instead, suppose \$I_b = 2 mA\$. Then \$\beta I_b = 200mA\$, which is much greater than \$I_c\$, so now the saturation assumption is correct.

Note: \$\beta I_b \gg I_c\$ is kind of a vague limit. Typically we use at least 10 times bigger for saturation.

For solving assuming the BJT is forward active, you would assume \$\beta I_b = I_c\$ and only this assumption. To check it (against saturation), simply make sure \$0.2V < V_{ce}\$.