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Just having some doubts. For the following circuit

enter image description here

No current to Vg

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    \$\begingroup\$ No, that's not correct. \$\endgroup\$ – Chu Oct 1 '18 at 22:07
  • \$\begingroup\$ With a closer look Vdd seems to be the only source. Vg and Vd are outputs. \$\endgroup\$ – Sparky256 Oct 1 '18 at 22:12
  • \$\begingroup\$ Shouldn't the formula first compute the voltage difference between Vdd and Vd and then scale: Vg = R2/(R1 + R2) * (Vdd - Vd)? \$\endgroup\$ – Alexander Pane Oct 1 '18 at 22:29
  • \$\begingroup\$ As you've drawn it, there's no current flowing through R1 or R2, so Vg = Vd = Vdd. Did you mean to have a voltage source on Vd or some other way of setting that node to a fixed voltage level? \$\endgroup\$ – brhans Oct 2 '18 at 3:16
  • \$\begingroup\$ Sorry. I should have clarified. No current to Vg but current can flow through Vd \$\endgroup\$ – AlfroJang80 Oct 2 '18 at 7:56
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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\$?

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My approach will be Vg = ((R2/(R1+R2))(Vdd-Vd)) + Vd

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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).

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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)).

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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.

schematic

simulate this circuit – Schematic created using CircuitLab

edit: You are right Alexander Pane, VDD was the wrong direction.

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    \$\begingroup\$ I really don't think that this is the same circuit that has been presented in the question. Also if the arrows indicate the voltage, they are in the opposite direction, if they indicate the current, then you have generators that consume power instead of generating power. So this type of circuit looks physically impossible. Check out the Passive sign convention: en.wikipedia.org/wiki/Passive_sign_convention \$\endgroup\$ – Alexander Pane Oct 30 '18 at 16:35

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