Superposition current with 2 loops:

[Color_of_Cyan]

Homework Statement

http://imageshack.us/a/img823/5397/homeworktest2prob4.jpg



Find the current i using superposition

Homework Equations

V= IR, KVL, KCL,

Superposition procedures

The Attempt at a Solution

I' + I'' = i

say I' then 7A source cut off

say I'' then 24V suppressed / silenced

For I': R tot = 5Ω

24V - 5Ω*I' - 3Ω*I' = 0

24V = 8Ω*I'

I' = 3A


but



if I do the second part (where 24V is silenced / suppressed), am I still looking at the current through the 3Ω again as I''?

If I try to do mesh equations with it, is the dependent source still called "3Ωi"?

I also got 0 for it anyways. What can I do about this?

http://imageshack.us/a/img22/3858/homeworktest2prob4edit.jpg

 

[gneill]

if I do the second part (where 24V is silenced / suppressed), am I still looking at the current through the 3Ω again as I''?

Yes.

If I try to do mesh equations with it, is the dependent source still called "3Ωi"?

Yes. Although you might find it simpler to write a node equation and solve for the potential at the top of the current supply, then find the current through the 3Ω resistor using it.

 

[Color_of_Cyan]

I ended up with

Vtop/3Ω + 7A = (-3Ωi + Vb)/2Ω

To end up with

-Vtop = -42V - 9Ωi

Then i plug into i'' = Vtop/3Ω

i'' = (42V + 9Ωi'')/3Ω

and get i'' = -7A but I don't think this is right.

and the current i would be -4A.Thanks though

 

[gneill]

Can you explain your node equation? What's Vtop and Vb? Is there a difference? What are the current directions implied by your equations?

 

[Color_of_Cyan]

I meant Vtop being the node above the 7A, and it was just supposed to be Vb, my bad

 

[NascentOxygen]

You can check that you are doing it correctly by comparing your answer with that obtained by another method. I find the voltage at the junction of the two resistors to be 29¼ volts, and from this you can determine the currents. Superposition should give the same result.

 

[The Electrician]

You can check that you are doing it correctly by comparing your answer with that obtained by another method. I find the voltage at the junction of the two resistors to be 29¼ volts, and from this you can determine the currents. Superposition should give the same result.

I get 20.25 volts at that junction.

 

[NascentOxygen]

EDITED: see later posts

[strike]

I get 20.25 volts at that junction. ✗[/size][/color]

∑ currents to that node must = 0.[/strike]

 

[The Electrician]

Are you getting 29.25 volts for the complete network, with all sources in place, or for the case where the 24 volt source is missing?

 

[gneill]

I concur with The Electrician on the node voltage when both independent sources are active: Vb = 81/4 = 20.25 V.

 

[NascentOxygen]

I concur with The Electrician on the node voltage when both independent sources are active: Vb = 81/4 = 20.25 V.

Very puzzling, as I already checked my working...

Ah! I think you are going to tell me that the 3i dependent source is not the current source I took it to be! http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon9.gif

Changing it to a dependent voltage source, I calculate 20¼ volts also.

 

[NascentOxygen]

I' + I'' = i

say I' then 7A source cut off

say I'' then 24V suppressed / silenced

For I': R tot = 5Ω

24V - 5Ω*I' - 3Ω*I' = 0

24V = 8Ω*I'

I' = 3A

Right.

if I do the second part (where 24V is silenced / suppressed), am I still looking at the current through the 3Ω again as I''?

If I try to do mesh equations with it, is the dependent source still called "3Ωi"?

It's a voltage of 3i”

I also got 0 for it anyways. What can I do about this?

Try again.
http://imageshack.us/a/img22/3858/homeworktest2prob4edit.jpg
i” + current source + current from dependent voltage source = 0

where current from dependent source

= (voltage across the 2Ω resistor)/2Ω

= ...

Can you fill in the dots here? We are aiming for only one unknown, viz., i”.

 

[Color_of_Cyan]

Still here.

Can't seem to get the math behind this i''Doing what nascent said above (and with the node junction above the 7A as "Vb":

Vb/3Ω + 7A + (3Ωi'' - Vb)/2Ω = 0

i'' = Vb/3Ω

Substituting the above into i'' above that would give 3Ω*(Vb/3Ω) as the whole coefficient in the equation so it would just cancel out as Vb - Vb = 0

Then i'' would end up as -3A and then i' + i'' would equal zero.

 

[gneill]

Still here.

Can't seem to get the math behind this i''Doing what nascent said above (and with the node junction above the 7A as "Vb":

Vb/3Ω + 7A + (3Ωi'' - Vb)/2Ω = 0

i'' = Vb/3Ω

Indicate on your circuit diagram the locations and directions of each of the current terms in the above equations. Do they agree with each other and the given information?

 

[NascentOxygen]

Doing what nascent said above (and with the node junction above the 7A as "Vb":
Vb/3Ω + 7A + (3Ωi'' - Vb)/2Ω = 0 ✗[/size][/color]
i'' = Vb/3Ω ✗[/size][/color]

You already have a direction for i” marked on your figure, so you are asking for trouble if you fail to adhere to that direction.

Also, you are confusing me (and probably yourself) by including the Ω symbol here. It reads as though the term Vb/3 has units of Ohms. It has units of Amperes, so you should omit the Ω. (If it were essential, you could make use of parentheses, e.g., Vb/(3Ω), but I see no desirability for this.)

Your diagram has directions already decided for the currents, so try redoing the above equations closely observing those directions.

 

[Color_of_Cyan]

Can you say that the i'' and 7A are going in and then the dependent source current is going out? Or are they all "going" in somehow?

For that one all 3 were going in.

I would get 0 for the i'' current so not sure.

I just can't seem to figure the right equations out for this part (which is probably because of that cursed dependent voltage source).

 

[gneill]

Can you say that the i'' and 7A are going in and then the dependent source current is going out? Or are they all "going" in somehow?

For that one all 3 were going in.

I would get 0 for the i'' current so not sure.

I just can't seem to figure the right equations out for this part (which is probably because of that cursed dependent voltage source).

I find it easiest to simply assume that all currents are entering or all currents are leaving a given node and write the equation accordingly; the math will sort itself out. That said, you must respect any fixed current sources and assign an appropriate sign. If the current source is flowing into the node and you choose to sum currents flowing in, the current will have a positive sign. If you choose to sum currents leaving, then it will have a negative sign.

So what would you like to do, sum the currents as entering or leaving? Choose one or the other (and it doesn't make any difference in the long run).

 

[NascentOxygen]

Can you say that the i'' and 7A are going in

That's how they are shown, so you need to tenaciously stick with that.

and then the dependent source current is going out? Or are they all "going" in somehow?

You get to choose. Mark in a current, and stay with that direction. Regardless, you can always write that the sum of the currents into the node = 0. (If any current has been marked as leaving the node, then it's equivalent to the negative of that current entering the node.)

The only way to get current i” through the 3Ω in the direction shown is for the left hand side to be considered more positive than the right, i.e.,

i” = (0 - Vb)/3

So try those equations again.

 

[Color_of_Cyan]

Ok, so how about:

Vb/3Ω + 7A = (-3Ωi'' + Vb)/2Ω

?

 

[NascentOxygen]

Ok, so how about:

Vb/3Ω + 7A = (-3Ωi'' + Vb)/2Ω

?

Are you representing KCL as―
• the current into a node = the current leaving?

But (Vb-0)/3 is the expression for current leaving the node via the 3 Ohm. Writing Vb-0 indicates that Vb is being considered the greater voltage (more positive) and it follows that the direction of the resultant current must be away from Vb towards 0. That direction corresponds to current leaving the node.

 

[Color_of_Cyan]

I meant to have Vb/3Ω as the current entering.

So it should be negative then?

(0 - Vb)/3Ω + 7A = (-3Ωi'' + Vb)/2Ω ?


I thought the potential over the 3Ω would be the same as only Vb so that is a little confusing (nodal problem).

It makes sense though if that's the way the current goes inward I guess.

 

[NascentOxygen]

That correctly represents current in = current out. So go ahead and solve it.

In circuit analysis, voltage direction (i.e., polarity of potential difference) and current direction are inextricably linked. Once you have nominated an arrow to represent one, you have by that action determined the arrow direction that must apply for the other.

 

[Color_of_Cyan]

Okay.

So for the second part with i'' and taking the top node as Vb:(0-Vb)/3Ω + 7A = (Vb - 3Ωi)/2Ωbut i'' = (0-Vb)/3Ω-Vb/3Ω -Vb/2Ω + 3Ωi''/2Ω = -7A-2Vb/6Ω - 3Vb/6Ω + 9Ωi''/6Ω = -7A(-5Vb + 9Ωi'')/6Ω = -7A

-5Vb + 9Ωi'' = -42V

-5Vb = 42V - 9Ωi''

Vb = (42/5)V + (9/5)Ωi''forgot to substitute i'' as

i'' = -Vb / 3Ω

so

Vb = (42V/5) + [ (9Ω/5)*(-Vb/3Ω) ]

Vb = (42/5)V - (3Vb/5)

(5/5)Vb + (3/5)Vb = (42/5)V

(8/5)Vb = (42/5)V

so

Vb = (21/4)Vi'' = -21V / (4*3)Ω

i'' = (-21/12)A

so

i = i' + i''

i = (36/12)A - (21/12)A

i = (15/12)A

i = (5/4)A is this correct?

 

[NascentOxygen]

so

i = i' + i''

i = (36/12)A - (21/12)A

i = (15/12)A

i = (5/4)A ✔[/size][/color]

is this correct?

These end results are the same as I calculated, though I haven't checked your working.