# Superposition theorem

Discussion in 'Homework Help' started by metelskiy, Nov 8, 2010.

1. ### metelskiy Thread Starter Member

Oct 22, 2010
66
3
i have this circuit

I got confused with positive and negative voltage supplies. In class we used only positive. I need to find voltage at point A using superposition theorem. Could someone please help me at least understand circuit? How would i redraw it in a more understandable way?
To start i know i have to short 12V source and leave -12V and calculate Voltage at A.
Than short -12v and leave 12v and calculate voltage at A again.
Than add those two values and it should be voltage at A for original circuit.
How would i treat that -12v souce? thanks.

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,452
707
You have two cases.

Case 1
12 V is shorted.
You have R1 parallel with R3, lets call it R1||R3.
The R1||R3 is in series with R2 and in series with -12 V source.
Draw this circuit, solve for what ever you need.

Case 2.
-12 V is shorted.
You have R2 parallel with R3, lets call it R2||R3.
The R2||R3 is in series with R1 and in series with 12 V source.
Draw this circuit, solve for what ever you need.

metelskiy likes this.
3. ### metelskiy Thread Starter Member

Oct 22, 2010
66
3
Thanks a lot. This circuit makes sense to me now. I found Va'=3.5V and Va''=-6.1V and Va=3.5v+(-6.1V)=-2.6V

Last edited: Nov 8, 2010
4. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
-2.6V matches my result.

hgmjr

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5. ### metelskiy Thread Starter Member

Oct 22, 2010
66
3
Ok here is another problem. I think i mainly understand it but if someone got time and patiance please check if i'm doing it right, first redrawing circuits. I'll need to find voltage at points A and B using superposition theorem.
Original (A and B on ends of R3):

5V and -12V shorted:

12V and -12V shorted:

12V and 5V removed:

6. ### Georacer Moderator

Nov 25, 2009
5,181
1,289
Your circuit look good to me. You just didn't label the nodes A and B in the initial circuit so I can't comment on the correct node notation.