Understanding how to decompose series and parallel circuits

[TheCanadian]

I have uploaded an image of the situation I am looking at. Essentially, the Thevenin voltage is being found, and one thing I don't quite understand in this example is the text saying R2 can be neglected when trying to find the Thevenin voltage. It says R1 and R3 form a voltage divider, but I don't quite see this since shouldn't the equivalent resistance between R2 and R3 be found in such a case, and this equivalent resistance acts as a voltage divider with R1? I found the Thevenin resistance, which was correct, and then applied $$ V_{Th} = I_{Short circuit}R_{Th} $$ but when I do this, I am slightly confused as to why when a short circuit is made, then R2 is neglected when trying to find the current through this area. Isn't the short circuit supposed to be directly between node A and B? Wouldn't this result in R2 still being included when finding $$ I_{Short circuit} \text{?}$$

Any insight into why R2 should be neglected and how $V_{Th}$ was computed would be greatly appreciated!

Edit: I seem to be clearly missing something. After looking at the example here, I found $R_{Th}$ correctly again, but don't quite understand why the numerator to find $$V_{Th}$$ is R2+R3...shouldn't it just be $$R_{Th}$$?

 

[tommyxu3]

Because in the first diagram, the current doesn't do through the resistance $R_2,$ so there is no $\Delta V$ between the point$A$ and the intersection on its left hand side.

 

[TheCanadian]

Because in the first diagram, the current doesn't do through the resistance $R_2,$ so there is no $\Delta V$ between the point$A$ and the intersection on its left hand side.

Thank you. But why is there a R3 in the numerator? Shouldn't that be R_Th?

 

[tommyxu3]

Because the whole circuit is like the series of $R_1$ and $R_3.$

 

[TheCanadian]

Because the whole circuit is like the series of $R_1$ and $R_3.$

You lost me there. Do you mind expanding on that point? If both are in series, why not include both R1 and R3 in the numerator?

 

[tommyxu3]

$\frac{V_1}{R_1+R_3}$ is the current of the circuit, and then through the resistance $R_3,$ it may take the voltage down $IR_3=\frac{V_1R_3}{R_1+R_3}.$

 

[TheCanadian]

$\frac{V_1}{R_1+R_3}$ is the current of the circuit, and then through the resistance $R_3,$ it may take the voltage down $IR_3=\frac{V_1R_3}{R_1+R_3}.$

Okay, so essentially, on the branch where R3 is located, we are trying to find the voltage through that position. My interpretation of the Thevelin voltage was that it should be the voltage of the original circuit measured at nodes A and B, so then why exactly are we finding it through another branch that is in parallel with this? It would make sense to me if R2 didn't exist, but I don't quite see how we can still neglect R2 in this case if we are simply saying the voltage through R3 will be equivalent to the voltage through nodes A and B.

 

[TheCanadian]

Any additional thoughts on why R3 is still included in the numerator?

 

[tommyxu3]

So what makes you confused is why $\Delta V$ isn't related to $R_2?$

 

[TheCanadian]

So what makes you confused is why $\Delta V$ isn't related to $R_2?$

Yes

 

[tommyxu3]

Mind that $R_2$ doesn't form a circuit, so the current will not pass it. Hence there's no $\Delta V$ between the both sides of it, right?