Dependent source with 2 other sources (using superposition), mesh

[Color_of_Cyan]

Homework Statement

http://imageshack.us/a/img834/5512/homeworkprobsg216.jpg Using superposition, find i.

Homework Equations

V = IR,

KVL, KCL,

Nodal / Mesh analysis,

voltage division, current divisionSuperposition procedure

The Attempt at a Solution

So this time the dependent source is a voltage source but there's a current relation, and all 3 are voltage sources. I know for superposition you're supposed to silence voltage sources 1 at a time and then find the current through each of those times and then add. But not sure what to do if there's a dependent source / 3 sources in a circuit.I think you silence the 2i since it's a voltage source so it's shorted so:

I' + i'' + I''' = 0

I think it goes:

I': 2v silenced
I'': 2i silenced
I''': 10V silenced

for i'' (2v silenced) and trying mesh:

i'' - i2 = 0 i think would be the mesh current down the 10V

10V - i''*20Ω - 2V = 0

and then 5Ω*i2 - 10V = 010V - i''*20Ω - 2V = 5Ωi2 - 10V

10V - i''*20Ω - 2V - 5Ω*i'' + 10V = 0

18V - 25Ω*i'' = 0

18V = 25Ω*I''

I'' = 0.72 AAny tips as to how I what I can do about that dependent source now? Is it possible to use mesh analysis with it? How? Or can I also use nodal?

 

[gneill]

For superposition, don't "silence" (suppress) dependent sources; only the independent sources are fair game. Keep one independent source alive at a time and sum the results.

You should take note that the 5Ω resistor plays no part in determining the current i, so that you can ignore it's loop. (why is that?)

 

[Color_of_Cyan]

So all the current goes just through the 10V source wire then?

 

[gneill]

So all the current goes just through the 10V source wire then?

Essentially, yes. The potential across an ideal voltage source cannot change, so it will always present the same potential change in any loop that it is part of. Nothing that happens on the 5Ω side of things can alter this fixed potential, so nothing to the right of the source can effect anything to the left of it. The 10V source effectively isolates the two loops from each other.

That leaves you with a single series circuit to deal with.

 

[Color_of_Cyan]

So what do I do about the dependent source now?:

I' + I'' = i

For I' (2V supressed):

10V = 3I'*20Ω = 0

I' = (1/6)Afor I'' (with the 10V supressed):

2V + 3I''20Ω = 0;

2V = -3I''*20Ω

I'' = (-1/30)A

(-1/30 + 1/6)A = (2/15)A but this isn't right...

 

[gneill]

So what do I do about the dependent source now?:

I' + I'' = i

For I' (2V supressed):

10V = 3I'*20Ω = 0

I' = (1/6)A


for I'' (with the 10V supressed):

2V + 3I''20Ω = 0;

2V = -3I''*20Ω

I'' = (-1/30)A

(-1/30 + 1/6)A = (2/15)A but this isn't right...

Sorry, I don't understand the equations that you're writing. Where dd the "3I's" come from? Isn't the dependent source a voltage source of value 2i?

 

[Color_of_Cyan]

Yeah, I added them in the same direction (but not sure if you can do that either)

 

[gneill]

Just write KVL for the loop. The dependent source has a voltage of "2i". Write it as "2i".

 

[Color_of_Cyan]

That's confusing, so it does NOT count as a current then? Even though it's "2i"? KVL would be a bit confusing here too.

So would they be:10V + 2i - i*20Ω = 0

and

2V + i*20Ω - 2i = 0 ?

 

[gneill]

That's confusing, so it does NOT count as a current then? Even though it's "2i"? KVL would be a bit confusing here too.

It's a voltage source. Think of it as supplying a voltage with a magnitude that's 2x that of the current i. It would also be fair to say that the '2' should have units of Ohms associated with it in order to keep things consistent in your equations.

So would they be:


10V + 2i - i*20Ω = 0

and

2V + i*20Ω - 2i = 0 ?

Yeah, that looks good. Be careful with units (see above). Solve for i in each of the equations.

 

[Color_of_Cyan]

Think of it as supplying a voltage with a magnitude that's 2x that of the current i. It would also be fair to say that the '2' should have units of Ohms associated with it in order to keep things consistent in your equations.

Wow... welp, probably no way I would've ever caught that then. I got for each current equation the first one

I' - (5/9)A

and the second one:

I'' = (-1/9)A

I' + I'' = I = (4/9)A and the answer is correctThanks.