Solve Vx using Superposition

In summary: V1 and V2 labels. Write KCL for both nodes. You'll end up with two equations in two unknowns. Solve for the one you want. are u saying that i should use node voltage analysis insted of this method.
  • #1
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Homework Statement



http://img16.imageshack.us/img16/8901/circuit1s.png [Broken]


Homework Equations


I need som help with this circuitI need to use superposition to find out the value of Vx.
And to do that i need to zero the current source and voltage source



The Attempt at a Solution


http://img694.imageshack.us/img694/2715/circuitug.png [Broken]

But i don't know what to do next because there isn't a single example in my book containing dependentsource.

Can anyone show me what to do.
I tried to find the current that flow through the 4 ohm resistance because if i can get a value of it then Vx=I*4.
But i don't know how to calculate that value.
I can't zero the 0,1Vx current source because it is dependent currentsource.
 
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  • #2
Presumably you are expected to use superposition to solve the problem, so you want to consider one supply at a time. As you surmised, you can't eliminate the dependent source.

So, suppose you first replace the voltage supply with a short. You should be able to write Kirchoff equations to find Vx. (I suggest using node current summation for the Vx node). Call the resulting value for Vx "Vx1".

Now replace the voltage supply and remove the current supply. Again solve for Vx, this time calling it Vx2.

Then use superposition to find Vx from the two previously determined values.
 
  • #3
Can you show me the KCL for the voltage supply
so i know how to do the same for current.
I have written KCL as are saying but i coudn't solve for Vx
 
  • #4
Perhaps you can show your attempt for the KCL? Then we can see what's going wrong and try to correct it.
 
  • #5
gneill said:
Perhaps you can show your attempt for the KCL? Then we can see what's going wrong and try to correct it.

KCL Picture 1
Current source=zero
KCL at node Vx in the diagram.
-0,1Vx+0,25Vx+(Vx-10)/20=0
(Vx in this equation is Vx1 as gneill said)
Kcl picture 2
voltage source=zero
Kcl
-2-0,1Vx+0,25Vx+Vx/20=0
(Vx in this equation is Vx2)
I haven't done any problem with dependent source using superpostion that's why i would really appreciate help.

Vx=Vx1+Vx2
 
  • #6
Your KCL equations look like they're fine. Did you try solving for Vx in each one?
 
  • #7
2,5 in the first one and 10 in the second one
total 12,5 V
Thanks Gneill.
Now i have a another problem that i need som help with. My teacher told me that i can zero the dependent source in this problem and the result that i got was correct but according to book and websites you can't zero dependent source.
so can anyone tell me if it is possible to solve this one without zeroing the dependent source. and How to solve it.
Pic of the circuit
I need to solve for V1 using superposition.
http://img98.imageshack.us/img98/2464/surajcircuit.png [Broken]

Uploaded with ImageShack.us
 
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  • #8
In this case there's only one independent source, the 1A current source, so superposition is sort of superfluous!

Consider: if you zero out the 1A current source, leaving the dependent source, then ix will go to zero and there will be no current anywhere in the circuit. The contribution to V1 will be zero. Why will ix go to zero? Because the only possible source for ix will come from the dependent source, and it only supplies 0.5 ix. It's a stable feedback situation that drives ix to zero.

Now remove the dependent source. You're left with a simple resistor network with a 1A source, and V1 can be solved easily. Now, the superposition contribution of the dependent source from above was zero, so this new value of V1 should be the total of the contributions. But this would mean that the dependent source has no effect at all on circuit behavior, which doesn't sound right.

So, if you leave the dependent source in the circuit you end up with two nodes to deal with, at the points with the V1 and V2 labels. Write KCL for both nodes. You'll end up with two equations in two unknowns. Solve for the one you want.
 
  • #9
So i shoudn't use superpostion in this case.
What do u meen by
if you leave the dependent source in the circuit you end up with two nodes to deal with, at the points with the V1 and V2 labels. Write KCL for both nodes. You'll end up with two equations in two unknowns. Solve for the one you want.
are u saying that i should use node voltage analysis insted of this method.


But how come my teachers way of solving this with superposition worked was it just a coincidence. I will ask him tomorrow.
 
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  • #10
surajalok said:
So i shoudn't use superpostion in this case.

Well, in my opinion you don't need superposition in this case because there's only a single "real" source that's producing current. If the dependent source had a constant factor in it (something like I = 0.5 Ix + 3A) then I'd tell a different story...

What do u meen by
if you leave the dependent source in the circuit you end up with two nodes to deal with, at the points with the V1 and V2 labels. Write KCL for both nodes. You'll end up with two equations in two unknowns. Solve for the one you want.
Should i use node voltage analysis

You can use whatever method you wish. You have two nodes, one with potential V1 and the other with potential V2. It should be a straightforwards matter to write an equation for each node that sums the currents entering each node to zero.

But how come my teachers way of solving this with superposition worked was it just a coincidence. I will ask him tomorrow.

I can't answer without knowing exactly what your teacher did.
 

What is superposition?

Superposition is a principle in physics that states that when two or more waves or fields interact, the resulting wave or field is the sum of the individual waves or fields.

How is superposition used to solve for Vx?

In the context of electrical circuits, superposition can be used to solve for Vx by breaking down a complex circuit into smaller, simpler circuits and analyzing them individually. The resulting voltages can then be added together to determine the total voltage at Vx.

What are the steps to solve for Vx using superposition?

The steps to solve for Vx using superposition are as follows:

  1. Identify all independent voltage or current sources in the circuit.
  2. Set all other sources to 0 (short-circuit for voltage sources, open-circuit for current sources).
  3. Calculate the voltage at Vx using the remaining sources.
  4. Repeat steps 2 and 3 for each independent source.
  5. Add all individual voltages together to determine the total voltage at Vx.

Are there any limitations to using superposition to solve for Vx?

Yes, there are a few limitations to using superposition. First, it only applies to linear circuits, meaning that the relationship between voltage and current must be proportional. Additionally, superposition does not take into account any nonlinearities or interactions between components in the circuit. Finally, it can be time-consuming and tedious to apply the principle to more complex circuits.

Can superposition be used to solve for other variables in a circuit?

Yes, superposition can be used to solve for other variables in a circuit, such as current or power, as long as the circuit is linear and the appropriate steps are followed. However, it may not always be the most efficient method for solving for these variables and other techniques may be more suitable.

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