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Hi, I'm trying to determine the value of Vo using superposition, however I've come across a hurdle when trying to zero the 9V voltage source. I tried simplifying the circuit using the following steps: enter image description here

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Up to this point, the Vo has remained consistent(2.077V), however upon combining the 2.25(2.3) and 1 ohm resistors in series, the value of Vo changes to 3. I understand why that is, as the resistor is in parallel with the 3V voltage source. What I don't understand is why Vo isn't 3 to begin with as I don't think I've simplified the circuit inaccurately. I would greatly appreciate it if someone could explain where I went wrong with this question. Thank you in advance

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    \$\begingroup\$ When you finally combined those last two resistor values, you destroyed Vo in the process. The node was removed when you series-combined the two remaining resistors. \$\endgroup\$ Commented Mar 24 at 3:09
  • \$\begingroup\$ Thanks for your reply. Could you explain what you meant by the Vo node being destroyed? Sorry, I'm new to this. Edit: to clarify I don't understand why the node was destroyed when I combined the two resistors in series, or why it wasn't destroyed when I combined the 3 and 9 ohm resistors in parallel \$\endgroup\$ Commented Mar 24 at 3:14
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    \$\begingroup\$ the node was in between the two resistors... you remove the in between \$\endgroup\$
    – jsotola
    Commented Mar 24 at 3:28

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After removing the 9V source, you have the upper circuit here, in which you combine R1 and R2 to be replaced by R5, underneath:

schematic

simulate this circuit – Schematic created using CircuitLab

Notice how node A becomes inaccessible after the replacement of R1 and R2 with a single component R5. It's now "inside" R5 somewhere.

Then you combine the parallel pair R5 and R3, but node B remains accessible. It is the left end of R4, which is still exposed even after the change:

schematic

simulate this circuit

Node B is at the junction of R4 and R6, which means when you combine them in your last step, B becomes inaccessible, existing somewhere "inside" the single resistor replacement:

schematic

simulate this circuit

In these examples, node B has potential \$V_O\$, which is your value of interest, and having "lost" it you can therefore no longer determine its potential. This last step was "one step too far", and you should return to the previous one, in which B was still accessible. It is a simple resistor potential divider, which may be easier to visualise like this:

schematic

simulate this circuit

Potential \$V_O\$ is:

$$ \begin{aligned} V_O &= V_1\frac{R_6}{R_4+R_6} \\ \\ &= 3V \times \frac{2.25\Omega}{2.25\Omega + 1\Omega} \\ \\ &= 3V \times \frac{2.25\Omega}{3.25\Omega} \\ \\ &= +2.077V \end{aligned} $$

When you said that \$V_O\$ changed to +3V in that final step, what you were actually measuring wasn't node B, it was node C, which was the only node still accessible after all those steps. Of course node C is now +3V; it always has been +3V, even before step 1, because it has been the positive terminal of voltage source \$V_1\$ throughout.

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