# Homework Help: Superposition for dependent sources and Pspice

1. Mar 5, 2013

### DODGEVIPER13

1. The problem statement, all variables and given/known data
The problem is uploaded it is problem number 2

2. Relevant equations

3. The attempt at a solution
I really just want confirmation that my work is ok, and the command for a dependent current source in PSpice. The things that I found that are dependent, are very confusing looking things instead of diamonds. I have tried E and F but I don't know how to use them?

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2. Mar 5, 2013

### Staff: Mentor

Check your work for problems 2,3, and 4.

I'm not sure about the syntax or format of the dependent current sources for PSpice; I make use of LTSpice where there are several versions of dependent sources.

3. Mar 5, 2013

### DODGEVIPER13

ok is there something wrong with them I will upload my work for 3 and 4

4. Mar 5, 2013

### DODGEVIPER13

Ok my work for 3 and 4 are uploaded

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5. Mar 5, 2013

### DODGEVIPER13

3 is on the left and 4 is on the right

6. Mar 5, 2013

7. Mar 5, 2013

### Staff: Mentor

For #2 I just note that at least one of the current values you've found doesn't agree with values I find. I don't recall having seen your detailed work on this problem.

For #3 it looks like you have a different answer this time, although I'm not sure if I'm interpreting your formatting correctly. What, as a number, do you get for the maximum current for Ix?

For #4, it looks like you're attempting a Δ-Y transformation (or two?). If so, the numbers don't look right to me. Can you describe the steps you're making?

8. Mar 5, 2013

### DODGEVIPER13

for number 2 that is one of ther first I uploaded. For number 3 I find for Ix' using the max power and resistance of each resistor that gives me an Ix' for each resistor. Then I find a Ix'' using voltage and total resistance. Which I then put into an Ix'+Ix''=Ix equation for both the 100 ohm and 25 ohm resistor. When I have done this I get 200mA for the 100 ohm and 100mA to 25 ohm. I then set up the inequalities to solve for the maximum current that should flow before ohmic heating destroys the limiting resistor. The values on the other side of the inequality are found from current dividers formula I25=Ix(Rseries/R100) and I100=Ix(Rseries/R25) the Ix given for I100 the current for the 100 ohm resistor and the 25 ohm resistor only using the Ix for 25 ohm resistor. So when I multiply .2(125/25) I get 1 A or 100mA so Ix<100mA and for the other I get 125 mA the lower number is the limiting so I choose it. For 4 R1=RaRb/(Ra+Rb+Rc), R2=(RbRc)/(Ra+Rb+Rc), and R3=RaRc/(Ra+Rb+Rc). I have marked what I am using for Ra Rb and Rc. I then converted the bottom two resistors 2 ohm and 3 ohm to 32/16 and 48/16 respectivley so that it would make since when I add the R1 and 2 ohm resistor together and the R2 and 3 ohm resistor together. When I do this I get 72/16 for the 2 ohm and R1. 63/16 for the 3 ohm and R2 resistors. The R3 goes on top of the 72/16 and 63/16 which are in parallel. I then put the R3 in series with the 72/16 and 63/16 resistances which gives me a total of 2.1A.

9. Mar 5, 2013

### DODGEVIPER13

For number for I am doing a delta-y transformation

10. Mar 5, 2013

### Staff: Mentor

Wow, a single massive paragraph to wade through...

If so I must have missed it. Sorry. Still, the results that you posted on the "answer sheet" don't match what I'm seeing when I solve the problem.
I don't follow what your Ix'' represents. When you apply an Ix to the circuit, both currents (through the resistors) change. This is because the node voltage where the resistors meet changes. Instead of looking at the maximum currents for each resistor, consider the maximum allowed potential drops across them. That will fix the maximum allowed potential at that node, from which you can then calculate the current Ix that it corresponds to.
I think you mean 2.1Ω. But that is not correct. Your method looks fine, but something went wrong in your reductions, perhaps in the determination of the parallel resistance.

11. Mar 5, 2013

### DODGEVIPER13

ok so for number 3 I use the .1A from before and mutiply that by 100 ohm to get 10 volts. and .2A also from before and muiply that by 25 ohm to get 5 volts. so the total potential at the node would be 5 volts because 10-5=5 volts. Am I good so far?

12. Mar 5, 2013

### Staff: Mentor

Not quite, but close. The maximum allowable drop across the 100Ω resistor is 10V, and across the 25Ω resistor is 5V. If the drop across the 100Ω resistor is 10V, that would mean the node potential would have to be 2.5V (since there's a 12.5V potential at the other end of the resistor, and 12.5V - 2.5V = 10V across the resistor).

So you have a choice of either 2.5V or 5V for the node voltage. The question then is, which one is produced by the maximum Ix. You should be able to work out Ix for both cases and choose the maximum. Consider applying nodal analysis to determine Ix, where in this case you're starting out knowing the node potential and Ix is the variable.

13. Mar 5, 2013

### DODGEVIPER13

For number 4 it should be 3.6 ohm correct?

14. Mar 5, 2013

### Staff: Mentor

Yes, that looks better

15. Mar 5, 2013

### DODGEVIPER13

Ok so doing the nodal analysis for 25 ohm Ix-(5/25)=0 so Ix=.2A. for the other 100 ohm resistor Ix-(10/100)=0 so Ix=.1A. I fear I have done the nodal analysis wrong! Should it be Ix-(2.5/100)=0 so Ix=.025A for the 100 ohm resistor?

16. Mar 5, 2013

### Staff: Mentor

You have to include the effect of both sources in your nodal analysis. Suppose Vx is the node voltage (which you'll set to either 2.5V or 5V later), then just write the node equation.

17. Mar 5, 2013

### DODGEVIPER13

Also I can re upload my work for problem 2 so I can get that fixed up as well?

18. Mar 5, 2013

### Staff: Mentor

Sure.

19. Mar 5, 2013

### DODGEVIPER13

Ok so I hope I am not irittating you with my ignorance but here goes my second attempt 2.5+100Ix+25Ix-12.5=0 so Ix=.08A for the 100 ohm. For the 25 ohm resistor 5+125Ix-12.5=0 so Ix=.06A. Once again I fear it is incorrect

20. Mar 5, 2013

### Staff: Mentor

Just write the node equation with variable Vx as the node voltage. Solve for Ix. Then you can plug in 2.5V or 5V for Vx and compare the values of Ix.

21. Mar 5, 2013

### DODGEVIPER13

Ok man do you have any more hints for me I dont see what I am doing wrong. I know that the voltage going in to a node is equal to the voltage going out.

22. Mar 5, 2013

### DODGEVIPER13

Ok nvm I think I found my error using my previous work it should be Ix < 125 mA

23. Mar 5, 2013

### DODGEVIPER13

However I would still like to proof using your method the reasons I dont get it maybe is because I am going by the books example which is exactly the same except with different numbers.

24. Mar 5, 2013

### Staff: Mentor

Current. The sum of the currents entering a node is equal to the sum of the currents leaving that node (or more traditionally, the sum of all currents entering and leaving a node is zero).

Have you written node equations before? It's essentially a matter of summing the expressions for the currents of the branches connected to the node by assuming the node voltages at each end of each branch.

25. Mar 5, 2013

### DODGEVIPER13

ok so on to number 2 because I feel as if I have been going at this too long I need to take a step back from it