# Universal Voltage Divider Bias Circuit with both BJT and JFET

I am dealing with the following circuit:

I am trying to understand if I can just assume that, much like in the JFET voltage divider circuit, V_G = 10/(40+10) * 16V , or if the presence of the BJT after the JFET will affect the voltage at this point.

Any help is appreciated. Thank you so much!!

The current into the base of the BJT is about $\beta$ (100) times smaller than the current through the emitter/collector. The upshot of this is that the effect of the 1k2 resistor as seen at the base can be regarded as the same as a grounded base resistor that is 100 times higher or 120kohm. This will make Vg a few percent lower.