Are My Superposition Circuit Equations Correct?

AI Thread Summary
The discussion revolves around solving a circuit using mesh current analysis and the superposition method. The original poster is confused about their approach, as their calculated currents all equal zero, indicating a mistake. Respondents clarify that the superposition method does not require closing sources; instead, they suggest keeping the sources active and applying Kirchhoff's Voltage Law (KVL) correctly. It is emphasized that careful definition of currents in the loops is crucial for accurate calculations. The conversation highlights the importance of understanding circuit analysis techniques to avoid common errors.
gomezfx
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Homework Statement


Find v0 and i0
[PLAIN]http://img214.imageshack.us/img214/2051/superquestion.jpg

Homework Equations


KVL and KCL

The Attempt at a Solution


I removed the voltage source and redrew the circuit with a short on the bottom left. I then used mesh current analysis with KVL and made a system of equations. My problem is that all of my current (i0, i1, i2, i3) come out to 0 which is definitely wrong.

Am I approaching this problem correctly and setting up my equations right?

my work:
http://img530.imageshack.us/img530/4779/superosition.jpg
 
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Looks like your on the right track! :)
 
With the 180 V source shorted, call the current through the 60 ohm resistor i1 (upward). Then the sum of the voltages around the loop on the lower left is 60i1 + 30(i1 + 7.5) = 0. This yields i1 = -2.5A straightaway. You can actually see that without the equations - clearly 2/3 of the current will go through the 30 ohm resistor and 1/3 through the 60. Looks like you forgot to use the 7.5 amps in that loop, and in the others. Very tough to find i0 . . .
 
To solve the circuit by super position method, you should not close the voltage source and open the current source. Keep them as it is and apply KVL and XCL.
 
Delphi51 said:
With the 180 V source shorted, call the current through the 60 ohm resistor i1 (upward). Then the sum of the voltages around the loop on the lower left is 60i1 + 30(i1 + 7.5) = 0. This yields i1 = -2.5A straightaway. You can actually see that without the equations - clearly 2/3 of the current will go through the 30 ohm resistor and 1/3 through the 60. Looks like you forgot to use the 7.5 amps in that loop, and in the others. Very tough to find i0 . . .

I'm having a hard time understanding this, is my circuit drawing correct? Because isn't the 60 not just i1?, isn't it 60(i1-i0)?
 
rl.bhat said:
To solve the circuit by super position method, you should not close the voltage source and open the current source. Keep them as it is and apply KVL and XCL.

Isn't that what the superposition method is though? Closing independent sources and solving circuits with only one source at a time.

So don't I need two separate circuits with only a current source or a voltage source?
 
gomezfx said:
Isn't that what the superposition method is though? Closing independent sources and solving circuits with only one source at a time.

So don't I need two separate circuits with only a current source or a voltage source?

Not necessarily. Superposition method involves asigning a current in each loop and applying KVL.
 
rl.bhat said:
Not necessarily. Superposition method involves asigning a current in each loop and applying KVL.

So I should just try solving the circuit as given originally?
 
You have to be extremely careful about defining i1. I defined it as the current through the 60 ohm resistor because I thought your i1 loop on the diagram was ambiguous.

There is an example of the superposition method here:
http://circuits.solved-problems.com/936/superposition-problem-voltage-current-sources/#more-936

Here is an article on doing it without superposition:
http://en.wikipedia.org/wiki/Mesh_analysis
I'm sure rl.bhat knows a great deal more about this than I do!
 
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