Superposition principle(circuits)

[kliker]

Homework Statement

i found this circuit somewhere here in this forum

https://www.physicsforums.com/attachment.php?attachmentid=10935&d=1190194219it asks to find vo

Homework Equations

The Attempt at a Solution

i found that vo is 6V using kirchhof's law, the current source i transformed it to voltage soure, and after some calculations i found that it's 6V

am i right?

but I can't find the 6 v using the superposition principle, can anyone help me out?

when i transformed to voltage source i said that 10ohm and 8 ohm resistors are in series? so the resistor of both is 18 ohm

then this 18 ohm resistor is in parallel with 6 ohm i found the new resistor and said that the current would be

I = 15/3 = 5 A

the voltage will be 5*(18||6+3)

then having the first source and making the same step i find another voltage then i add them up and I don't get 6 V to be the Vo

 

[Antiphon]

First find the voltage contributed by the current source by making the voltage source a short circuit.

Then find the voltage contributed by the voltage source by making the current source open-circuit.

Add the voltages contributed to the resistor from each.

 

[kliker]

First find the voltage contributed by the current source by making the voltage source a short circuit.

Then find the voltage contributed by the voltage source by making the current source open-circuit.

Add the voltages contributed to the resistor from each.

you mean something like this?

[PLAIN]http://img405.imageshack.us/img405/4199/91234315.jpg

 

[Antiphon]

No. You draw two different schematics.

In the first one, there is no current generator you just leave it off. That's what it means for it to be "open circuit".

In the second one, you have the current generator but you replace the voltage generator with a wire. Just a plain wire, like it's not there. That's what it means to short it out.

You solve the two problems and get two voltages. You add them up and that would be the voltage on the resistor in the original circuit. That's what superposition means, you add solutions.

A current generator with zero current is an open circuit (it's not there.)
A voltage generator with zero volts is just a wire. It's there but zero volts.

 

[kliker]

No. You draw two different schematics.

In the first one, there is no current generator you just leave it off. That's what it means for it to be "open circuit".

In the second one, you have the current generator but you replace the voltage generator with a wire. Just a plain wire, like it's not there. That's what it means to short it out.

You solve the two problems and get two voltages. You add them up and that would be the voltage on the resistor in the original circuit. That's what superposition means, you add solutions.

A current generator with zero current is an open circuit (it's not there.)
A voltage generator with zero volts is just a wire. It's there but zero volts.

i can't really understand 100% the whole procedure, so for the first source i should have this

[PLAIN]http://img269.imageshack.us/img269/5896/ph1q.jpg

here Itot = E /Rtot

but what's the Rtot? 24 Ohms or 14?

if we have Itot

V1 = Itot*Rtot

then for the second one we have

[PLAIN]http://img263.imageshack.us/img263/2521/ph2r.jpg

here Itot = 15/Rtot

and Rtot is 9

so V2 = Itot*9

then Vtot = V1+V2

?

Our teacher didnt explain this method at all, the book has no explanation and I am having a bad time with it

i can't understand why we should use this principle when there is kirchhoff rules

using kirchhof rules i get 6 V is this result correct?

 

[Antiphon]

No, not quite. When you replace the voltage generator with a short, you still need the resistor connected to it. That part of the circuit is there, its only the voltage source that gets replaced by the wire.

The second circuit you drew is ok because when the current generator goes open circuit, you no longer flow current through the resistor on the left.

Your first post said you can't get the answer using the superposition principle. I am using the superposition principle for the sources, calculating the contribution of each source one at a time. You should use the method that makes the most sense to you.

 

[kliker]

No, not quite. When you replace the voltage generator with a short, you still need the resistor connected to it. That part of the circuit is there, its only the voltage source that gets replaced by the wire.

The second circuit you drew is ok because when the current generator goes open circuit, you no longer flow current through the resistor on the left.

Your first post said you can't get the answer using the superposition principle. I am using the superposition principle for the sources, calculating the contribution of each source one at a time. You should use the method that makes the most sense to you.

thanks for your help i understand it now

one last question

when we have a current source like this

https://www.physicsforums.com/attachment.php?attachmentid=10935&d=1190194219

the one at the beginning and we want to make it voltage source
V = 3*10 = 30 Volt right?

but when we make it to voltage source we will have this

[PLAIN]http://img710.imageshack.us/img710/2458/87121027.jpg

now about this E = 30

is this the voltage that the circuit element gives to the circuit with the resistance of 10 Ohm

or is it just the voltage that it gives to the circuit without the resistance of 10 ohm?

if it is the second then the voltage that it gives will be 30 - Itot*10 right?

but i don't think it's the second what do you think?

 

[Antiphon]

It's fully equivalent in all cases. All you have to do is check these two: do you get 30 volts from both? Yes.

Do you get 3 amps from both if you short them? Yes you do because 30 volts onto 10 ohms gives three amps.

You can still use superposition of sources in your new circuit. You'd replace one of the voltage sources with a wire and compute using the other source, then vice versa. Ad the contributions just like before. You should get the same answer.

 

[kliker]

thanks a lot for your help :)