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I am supposed to solve (to find voltage and current on resistor R2) this circuit using Thévenin's theorem. How did we get the formula for open circuit voltage? And why are we treating resistances as conductances?

enter image description here

Moreover, I don't seem to be able to find generally applicable way to calculate Ui. I'm really new to circuits and the formulas for Ui seem to be significantly different for each different circuit.

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  • \$\begingroup\$ Why are you using "U" instead of "V" or "E" as symbols denoting voltage difference? \$\endgroup\$ – EM Fields Oct 24 '16 at 22:21
  • \$\begingroup\$ @EM Fields It is written by our university lecturer, it's an actual standard in this country. \$\endgroup\$ – John Smith Oct 24 '16 at 22:42
  • \$\begingroup\$ electronics.stackexchange.com/questions/99584/… \$\endgroup\$ – jonk Oct 24 '16 at 23:32
  • \$\begingroup\$ "I don't seem to be able to find generally applicable way"? Really, how about Ohm's Law?! Vth = (R3)I where I = R3(U/(R3+R4)). \$\endgroup\$ – CroCo Oct 25 '16 at 1:48
  • \$\begingroup\$ @EMFields In the Netherlands we learned to use "U". to indicate a potential difference. Original german "Unterschied" meaning difference. \$\endgroup\$ – Decapod Oct 25 '16 at 9:59
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Remove \$R_2\$, to give an open circuit in order to determine \$V_{TH}\$.

\$R_3\$ and \$R_4\$ then form a voltage divider across \$U\$, and \$U_2= V_{TH}\$.

Thus, \$V_{TH}=\dfrac{R_3}{R_3 +R_4}U\$. I guess \$U=10 \:V\$.

You can change the resistances into conductances to get the formula in your question, but there's no need to do this. Some prefer to work in G, but mixing the two in one formula is really confusing. Perhaps that was the aim of the question.

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I don't have the ability to comment yet, but there is some missing information here? This isn't a complete question. I think this is all correct...

Note that resistance and conductance are just reciprocals of the same quantity... $$\frac{1}{R_N} = G_N \textbf{ and } \frac{1}{G_N} = R_N$$

Anyway, let's imagine redrawing the circuit as follows just by disconnecting \$R_2\$, which we can do because this is essentially going to be our Thevenin terminals, which we shall open-circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

We want to find the voltage across and current going through \$R_2\$ (in the original circuit, not the one where \$R_2\$ is disconnected, obviously). So, we need to find a Thevenin equivalent for everything on the right-hand side of the above figure...

So, let's start by finding a Thevenin resistance, \$r\$:

In order to do this, we must make \$V\$ a short-circuit, leaving us with:

schematic

simulate this circuit

Now (hopefully, you can find equivelant resistances for simple parallel/series networks?):

$$r_{th} = \frac{R_4 \times R_3}{R_4 + R_3} + R_1$$

We also need to find a Thevenin voltage, \$V_{th}\$:

schematic

simulate this circuit

Let's name the nodes as above, such that:

$$V_{th} = V_3 - V_1 = V_3 - V_2 = R_3 \times I $$

Where \$I\$ is the current flowing through \$R_4\$ and \$R_3\$.

Now we need an expression for \$I\$ (obtainable using KVL, which you're OK with, hopfeully?):

$$V - R_4 I - R_3 I = 0$$ $$V - I(R_4 + R_3) = 0$$ $$I = \frac{V}{R_3 + R_4}$$ $$\therefore V_{th} = \frac{V \times R_3}{R_4 + R_3}$$

schematic

simulate this circuit

Now we have what we want:

$$I_{th} = I_2 = \frac{V_{th}}{R_2 + r}$$ $$V_2 = R_2 \times I_{th}$$

You should check out this webpage for more detailed explanations... and do some Googling. There isn't really a general 'formula' to work out \$V\$... otherwise there would be no 'analysis' in circuit analysis.

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  • \$\begingroup\$ I think Vth = (R3) I where I = U/(R3 + R4). \$\endgroup\$ – CroCo Oct 25 '16 at 1:41
  • \$\begingroup\$ @CroCo didn't spot that. Thanks! Updated... \$\endgroup\$ – jmdlok Oct 25 '16 at 16:16

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