if someone wouldn't mind verifying my solution to this problem by superposition?

http://i615.photobucket.com/albums/t...ionproblem.jpg

It took me a while but I think I got it.

I used KVL for the voltage source and got -12 -3i -3i = 0 so i' = 2Amps

for the Current source I used current division and got both currents to equal 1.5Amps so i'' = 1.5Amps going the other direction

When adding them both together since i'' was going the other way against i' yielded i'-i'' = 0.5 Amps flowing to the right. so i = 0.5 A. Is that right? I can't see why not but I spent so much time and confusion I can never be sure. Is it right?

 

Hello chaz0426,
Yes you are right

 

arly an ideal current source Hi chaz0426
I think on the voltages source as well on the current source you need to have the resistance
too. If you suggest a ideal votlage source the resistance would be 0 and there is always
12 V on it. Similarly the current source would deliver the current of 3 A at any voltage across.
First the 6 Ohm Resistor is just a load to the voltage source.
The solution is a current of 0,5 A from right to left leading to a voltage of 10.5 V at the
current source and a current of 3.5 A throug the vertical 3 Ohms.

Regards Christian