# The superposition theorem with dependent sources

1. Feb 26, 2013

### DODGEVIPER13

1. The problem statement, all variables and given/known data
Use the superposition principle to determine the value of Ix.

2. Relevant equations
I1+I2+I3=0 KCL
V1+V2+V3=0 KVL
Ix'+Ix''= answer
3. The attempt at a solution
So I understand how to get Ix' its just standard KVL it is -10+2Ix'+Ix'+2Ix'=0 which solving for Ix' gives 2A. This is the part I dont get so using nodal analysis (V''/2)+((V''-2Ix'')/1)=3 which when I simplify I get 6=3V''-4Ix''. So staying on this part why V''/2 is it this because that gives the current that goes across the 2 ohm resistor and V'' is the voltage on the wire? Also the second part ((V''-2Ix'')/1) is it this because your getting the current going from left to right over the 1 ohm resistor and thus by using the potential over 1 ohm which gives the current flowing from the left side of the resistor to the right. You would subtract off the dependent source part because it flows against the other current, so you get the current that was being solved for which is from left to right? essentially you are doing this I1=I2+I3 where I1=3A. So am I right there? On the next part I am confused Vx''+2Ix''=0 which when solved gives Vx''=-2Ix'' is this because the 3A source is in parallel with the dependent source and so the voltage is the same across both and thus V''=-2Ix''. Or is it because of KVl and so starting from the far left of the circuit I get 2Ix''+Vx''=0. I guess what I am asking is am I doing so form of KVL here or is it just simply relating the dependent source quantity. Anyways when this is all done I get the answer which is 2A.

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2. Feb 26, 2013

### DODGEVIPER13

Sorry I meant 1.4 not 2A on the last part

3. Feb 26, 2013

### DODGEVIPER13

Sorry if it is really confusing

4. Feb 27, 2013

### DODGEVIPER13

Oh also is my picture readable im sorry if it is too small let me know

5. Feb 27, 2013

### DODGEVIPER13

I can re upload or clarify anything just lemme know please

6. Feb 27, 2013

### Staff: Mentor

7. Feb 27, 2013

### DODGEVIPER13

Ok thanks on his problem I get Ix''=-1.75 A is that correct

8. Feb 27, 2013

### DODGEVIPER13

So 3-1.75=1.25

9. Feb 27, 2013

### DODGEVIPER13

Where v''=-3Ix'' but why is this is it because of KVL equations giving me 3Ix''+v''=0 or is it just simply a relation

10. Feb 27, 2013

### Staff: Mentor

Once you have worked out a value for the current, use Kirchoff's Laws to confirm that that value satisfies ∑ currents into a node = 0. If it does, it must be right. Or check your current together with the knowledge that the voltage at the centre node is 20¼ volts.

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