I am not sure how to attempt this question as Vin isnt given. If it was I would:

*Work out the first Vout after the current has gone through 1kΩ
*I would use the first Vout to work out the final Vout after the current has gone through the second 1kΩ resistor
*For these calculations I would use the voltage division formula:
Vout = Vin x R2 / R1 + R2.

How would I work out Vin, in order to do the calculations above?

 

Well assume that Vin=1V

Or we can write nodal equation an solve unknowns.

\(\frac{(Vin-Va)}{R2}=\frac{Va}{R4}+\frac{(Va-Vout)}{R2}\)

\(\frac{(Va-Vout)}{R2}=\frac{Vout}{R1}\)

But Va can be found by inspection (because we have here two voltage divider)
First
Va = ( R4||(R2+R1)/ ( R2 + R4||(R2+R1) ) * Vin

\( Va=Vin *\frac{\frac{R4*(R2+R1)}{R1+R2+R4}}{R2+\frac{R4*(R2+R1)}{R1+R2+R4}}\)

second

Vout = Va * R1/(R2+R1)

 

Oh ok I see. Something else I am curious about is how the various currents flow in this circuit. Would it be like this:

*Current starts flowing at Vin (I1)
*It then goes through R2, splits at Va (the one goes through R4 to earth (I2), and the other goes through R2 (I3) therefore I1 = I2 + I3 )
*At the next common point, spilts again (one goes through R1 to earth (I4), and the other one goes through Vout (I5) (I3 = I4 + I5).

 

*At the next common point, spilts again (one goes through R1 to earth (I4), and the other one goes through Vout (I5) (I3 = I4 + I5).

You have to assume that no current flows out of Vout. Imagine it as a test probe point. If current flowed out and we had no way of knowing what it was, the system would be unsolvable.

 

Here is the Millman's Theorem analysis technique applied to the circuit.

I have co-opted jony130's diagram to permit comparison between the two approaches.

Begin by forming (by inspection) the Millman's Theorem equation for Vout in terms of Va:

\(V_{out} \,=\, \Large \frac{\frac{V_{a}}{R_2}}{\frac{1}{R_2}+\frac{1}{R_1}}\normalsize \, \, (1)\)

\(V_{out} \,=\, \frac{R_1V_{a}}{R_1+R_2}\normalsize \, \, \, \, (2)\)

Rearrange equation 2 to isolate Va on the left side of the equal sign:

\(V_{a} \,=\, \frac{(R_1+R_2)V_{out \.}}{R_1}\normalsize \, \, (3)\)

Next, form the Millman's Theorem equation (by inspection) for Va in terms of Vin:

\(V_{a} \,=\, \Large \frac{\frac{V_{in}}{R_2}}{\frac{1}{R_2}+\frac{1}{R_4}+\frac{1}{(R_2+R_1)}}\normalsize \, \, \, \, (4)\)


\(V_{a} \,=\, \large \frac{R_4(R_2+R_1)V_{in}}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4}\normalsize \, \, (5)\)

Set equation (3) equal to equation (5) and solve for Vout/Vin:

\(\frac{(R_1+R_2)V_{out}}{R_1} \,=\, \large \frac{R_4(R_2+R_1)V_{in}}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4}\normalsize \, \, (6)\)

\(\frac{V_{out}}{R_1} \,=\, \large \frac{R_4V_{in}}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4}\normalsize \, \, (7)\)

\(\frac{V_{out}}{V_{in}} \,=\, \large \frac{R_1R_4}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4} \normalsize \, \, (8)\)

Done.

hgmjr