The B-E junction looks like a diode in its V vs I curve (silicon diode) -
So when the switch is initially closed, the diode is not in conduction, so
its an infinite R, so that "R" is a V divider with the 10 K and essentially all
10V is applied to the junction, eg. no current flow so no drop across 10K.
Then thru diode field theory/operation it starts to conduct and the following
occurs ( V settles to Vfwdjunction, ~ .7 silicon, ~.3 germanium).
Video on barrier V and basic operation -
The actual physics a tad complicated, lots of delicious field theory every EE loves
on exams......
but that is the seat of the pants description.
That circuit by the way is maybe a possible problem. When you want a transistor to
operate as a switch, to change its Vce to as low as possible when on, there is a rule
of thumb used as to how much current goes into the base. Its called "forced beta",
and basically one wants to force into the B-E junction Ic / 10 minimum. That will
generally saturate the transistor. If you dont and the Vce is significant you get a lot
of power dissipated in transistor, could burn it out if not heat sunk well enough or
if the transistor geometry is not adequate to handle the power or if its thermal R
is not low enough to get the heat out of the junction/package.
Still a pretty good resource is this book -
Regards, Dana.