Superposition – another

It seems to be a problem for students, so here is another example:

Superposition circuit to start with
Superposition circuit to start with
Using superposition Theorem, it will be analysed without the second voltage source.
Starting with first voltage source
Starting with first voltage source
 Analyse Vab due to voltage source V1(36V). The total resistance RT=6+4//(4+8) (visually)=6+4×12/16=6+3=9. The current IT=36V/9=4A.
The voltage dropped across R1=6×4=24V, so VR2=36-24=12V which is VR3+VR4. Using voltage divider to find Vab=VR4=12×8/12=8V.
Next is to find Vab due to V2(48V).
Now due to the second voltage source
Now due to the second voltage source
Same as before, RT=4+6//(4+8)=4+6×12/(6+12)=4+4=8 ohms. IT=-48V/8 ohms=-6A.

The the drop of R2=(-6)x4=-24V and VR1=-48-(-24)=-24V which is the same as VR3+VR4 because they are in parallel. Using voltage divider again to have Vab=VR4=-24×8/12=-16V.

Final step

Combining the two (superposition theorem says): Vab=8V-16V=-8V.

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