Just having some doubts. For the following circuit
No current to Vg
No, that's not correct. Here are some hints.
First, what is the total voltage across R1 and R2?
Now, what is the current flowing through R1 and R2?
What is the voltage across R2 alone?
The bottom end of R2 is at \$V_D\$. The top end of R2 is your \$V_G\$. How is \$V_G\$ related to the voltage across R2 and \$V_D\$?
My approach will be Vg = ((R2/(R1+R2))(Vdd-Vd)) + Vd
I would recommend you to find the drop across R1 and the remaining is the drop across R2 and Vd.. V(R1)=R1*Vdd/(R1+R2)....this is the drop across R1
Remaining voltage is Vdd-V(R1) and this is equal to V(R2)+Vd.
Since V6 is in parallel, Vg=Vdd-V(R1).
You can break up the problem in some mini steps:
As you said, there is no current in Vg and I also suppose that Vd is not on an open-circuit, otherwise there would be no current and all endpoints wouls have a Vdd potential, but I don't think that is the case.
If all the current flows from Vdd to Vd, this means that the current in the resistors R1 and R2 (lets call the currents I(R1) and I(R2)) are equal, so I(R1) = I(R2) = I.
for Ohm's law you have:
Vg - Vd = I x R2, and also:
I x R1 + I x R2 = Vdd - Vd. You can express this last one for I:
I = (Vdd - Vd) / (R1 + R2).
You can substitute the I of the first expression with the latter one:
Vg - Vd = (Vdd - Vd) x (R2 / (R1 + R2)).
If Vd is your reference (0V in your reference system), you would have:
Vg = Vdd x (R2 / (R1 + R2)).
It is always helpful to redraw a schematic in order to see better what is going on. Vd is just an unknown potential or "voltage referenced to GND". In this case the KVL and KCL (Kirchhoff's laws) will lead you to the solution.
Vg = Vd+I_R2*R2 and I_R2 = Vdd-Vd/(R1+R2) so Vg = Vd + (Vdd-Vd)/(R1+R2) * R2
Now assume Vd = 0 and you will end up with the "well-known" voltage divider formula.
edit: You are right Alexander Pane, VDD was the wrong direction.