Ok so in voltage divider biasing of base of NPN BJT transistor, the voltage divider determines the quiescent voltage. 10k and 10k would put 4.5v across the base. 100 ohm and 100 ohm for R(L) and R(E). Doesn't current flow from the bias resistors at the base to ground through the emitter? How come the voltage is still 4.5v (-the voltage drop)? Wouldn't the emitter be practically a short circuit compared to the bottom resistor in the divider/bias section? Is there any voltage divider circuit that works without an emitter resistor causing negative feedback?