Use superposition to find potential drop (but a problem)

[Color_of_Cyan]

Homework Statement

Use superposition to find V₀ in the circuit:


http://imageshack.us/a/img62/5567/eecircfinal1.jpg

Homework Equations

V = IR

Voltage division for 2 series resistors:
V across a resistor = (Resistor / total series resistance)(V in)

Current division for 2 parallel resistors:
Current through a resistor = (other resistor / math. sum of parallel resistors)(I in)


Superposition methods:
I = I' + I''

I' = current with circuit having voltage source shorted
I'' = current with circuit having current source cut out

The Attempt at a Solution

Following the above & So for I' and the voltage source shorted:

http://imageshack.us/a/img29/8991/eecircfinal1edit1.jpg


Then with current division,

I' = (4Ω/12Ω)(6A)

I' = 2A


Following the above again for I'' and with the current source cut out:

http://imageshack.us/a/img541/5840/eecircfinal1edit2.jpg


And then since the resistors above are all in series then I just add to get R tot = 12Ω.

But then I = V/R

So then I'' = 24V / 12Ω

I'' = 2A,

BUT compared with the set superposition current direction in the diagram would it then be -2A?


If so, then the current I = I' + I'' would be 0 and then I think that would be wrong. I know afterwards after you get I from I' + I'', then V₀ is just V₀ = I*2Ω

 

[gneill]

BUT compared with the set superposition current direction in the diagram would it then be -2A?

Sure.

If so, then the current I = I' + I'' would be 0 and then I think that would be wrong. I know afterwards after you get I from I' + I'', then V₀ is just V₀ = I*2Ω

0 is a perfectly good value for the current

Why not confirm your results using one or more other techniques? For example, nodal analysis, or mesh analysis. Or even better, source conversion; convert the voltage source and series resistor into an equivalent current source... what do you see then?

 

[Color_of_Cyan]

Okay, so it's a trick question then :D

Thanks.

So V₀ is just 0 then too.

And checking what you say, switched the Thevenin voltage to Norton equivalent current (or however you call it) and got 6A going the opposite way. So the current actually cancels out ?
Thanks for the help again.

 

[gneill]

And checking what you say, switched the Thevenin voltage to Norton equivalent current (or however you call it) and got 6A going the opposite way. So the current actually cancels out ?

The current sources are in parallel, so they add: -6 + 6 = 0, no problem!