Solving Mesh Currents: Find 3 I's & 5A Current

[Color_of_Cyan]

Homework Statement

http://imageshack.us/a/img90/4018/homeworkprobsg25.jpg

Find the three mesh currents

Homework Equations

V = IR,

KCL, KVL, Mesh / maybe nodal

The Attempt at a Solution

I'm not quite sure how to set up the equations

but I#1 = 4A right off the bat, for sure. I may need some more work with KVL.So for I#2 (mesh 2):

10V - (1Ω)(I₂) = 0For I3 (mesh 3):

(1Ω)(I#3) + (I#3)(2Ω) = 0For I#1 (mesh 1) even though I know what I#1 is:

(I#1 - I#3)1Ω + (I#1 - I#2)1Ω = 0Would these be correct equations?

Not sure how the 5A current going in would fit in here either.

 

[gneill]

There will be some unknown voltage drop across the 5A current source, so introduce a variable for it and include it in your mesh equations for loops 2 and 3. The 5A supply also introduces a constraint equation relating I2 and I3, giving you the additional equation needed because you've introduced another variable.

 

[Color_of_Cyan]

Sorry I forgot, but in the circuit image the bottom node junction right under the 5A was marked 'b'.

So, can I say

10V - (1Ω)(I#2) - (Va - Vb) = 0

and then

(Vb - Va) - (1Ω)(I#3) - (2Ω)(I#3) = 0 ??

And for the current, can I say:

I#2 - I#3 = 5A

 

[gneill]

Sorry I forgot, but in the circuit image the bottom node junction right under the 5A was marked 'b'.

So, can I say

10V - (1Ω)(I#2) - (Va - Vb) = 0

and then

(Vb - Va) - (1Ω)(I#3) - (2Ω)(I#3) = 0 ??

And for the current, can I say:

I#2 - I#3 = 5A

Sure, but why introduce two variables when one will do? Let Vx = (Va - Vb). Use Vx.

 

[Color_of_Cyan]

Okay, I thought you were supposed to do it that way. It tricked me into thinking nodal.

I still got the wrong answer though, doing it by:

10V - (I2)(1Ω) - Vx = 0

I2 - I3 = 5A , and lastL

Vx - (I3)(1Ω) - (I3)(2Ω) = 0, simplified to:

Vx - (I3)(1Ω) - (I3)(2Ω) = 0Then I2 = 5A + I3

Plugging I2 into last equation gives

-Vx - 5V - (1Ω)(I3) = 0, moving the 5V to the other side:

-Vx - (1Ω)(I3) = 5V, and lastly there's still the equation:
Vx - (3Ω)(I3) = 0

combining give:

5V-(4Ω)(I3) = 0

then I3 = (5/4)A but that's not right

 

[gneill]

Looks like you've forgotten to include the effects of known current I1 in your equations.

 

[Color_of_Cyan]

May I ask how you would go about expressing that for this circuit? I can't see how.

 

[gneill]

May I ask how you would go about expressing that for this circuit? I can't see how.

I1 (the known 4A mesh current) flows in the opposite direction to I2 and I3 in the resistors...

 

[Color_of_Cyan]

So you're saying do KVL for mesh 1 too?

Then:

(I1 - I3)1Ω + (I1 - I2)1Ω = 0 ?

(and then substitute 4A = I1)So are these correct along again with:

I2 - I3 = 5A , and lastL

Vx - (I3)(1Ω) - (I3)(2Ω) = 0, simplified to:

Vx - (I3)(1Ω) - (I3)(2Ω) = 0What could you say could be a way to tell + and - when doing KVL like this if there is something wrong.

 

[gneill]

So you're saying do KVL for mesh 1 too?

No, but you do have to take into account the fact that mesh current I1 = 4A is flowing through the two 1Ω resistors when you write the equations for loops 2 and 3.

<snip>

So are these correct along again with:

I2 - I3 = 5A , and lastL

Vx - (I3)(1Ω) - (I3)(2Ω) = 0, simplified to:

Vx - (I3)(1Ω) - (I3)(2Ω) = 0

I'm not seeing the effect if I1 there.

What could you say could be a way to tell + and - when doing KVL like this if there is something wrong.

Usually you'll get wrong results Plug the results into the circuit and confirm KVL and KCL consistency.

 

[Color_of_Cyan]

So

I2 - I3 = 5A

Vx - (I3 - 4A)(1Ω) - (I3)(2Ω) = 0

10V - (I2 - 4A)(1Ω) - Vx = 0

 

[gneill]

So

I2 - I3 = 5A

Vx - (I3 - 4A)(1Ω) - (I3)(2Ω) = 0

10V - (I2 - 4A)(1Ω) - Vx = 0

Yeah, that's looking better

 

[Color_of_Cyan]

and I get the exact same wrong answer for I3 still -_- :I2 = 5A - I3

Vx - (I3 - 4A)(1Ω) - (I3)(2Ω) = 0

10V - (I2 - 4A)(1Ω) - Vx = 0Then:

Vx - 1ΩI3 - 4V - I32Ω = 0

10V - I21Ω + 4V - Vx = 0Merging the two above:

10V - 1ΩI3 - 1ΩI2 - 2ΩI3 = 0, simplifies to:

10V - 3ΩI3 - 1ΩI2 = 0 Plugging I2 = 5A + I3 into the above gives:

10V - 3ΩI3 - 1ΩI3 - 5V = 0

5v = 4ΩI3

I3 = (5/4)A again which is wrong. The correct answer it says is I3 = (13/4)A and I guess I still don't know...

am I still missing something?

 

[gneill]

I think something went wrong with your algebra when you merged the two mesh equations. How did you go about it? Solve the first for Vx and substitute it into the second? If so, perhaps a sign issue...

 

[Color_of_Cyan]

What would be the best way to deal with the sign issue though? That's why I ask, how can I better tell which is positive and which is negative... I need work with KVL in regards to that, unfortunately. Trying what you said:

Vx = 1ΩI3 + 4V + I32Ω

10V -1ΩI2 + 4V - Vx = 0, so

10V -1ΩI2 + 4V - 1ΩI3 - 4V - 2ΩI3 = 0So again: I2 = 5A - I3

10V - 5V - 1ΩI3 + 4V - 1ΩI3 - 4V - 2ΩI3 = 0

5V - 4ΩI3 = 0 , same thing

 

[gneill]

Why aren't you gathering the like terms in your expressions? For the Vx equation you have two terms with I3. In the next equation, two separate voltages. Why not simplify your life by simplifying these before proceeding?

Dropping the units for now,

Vx = 3*I3 - 4
14 - I2 - Vx = 0
I2 = 5 + I3

Note that last equation.

Regarding the signs of the terms, they are specified when you draw in your mesh currents. The direction of the mesh current tells you the assumed direction of potential drops through resistances, and whether there is a potential drop or rise when going through a voltage supply in the direction of the current.

When you "walk around" the loop in the direction of the mesh current, potential will drop when passing over a resistance. If a neighboring loop current goes in the opposite direction then it will cause a potential rise in a shared resistance.

 

[Color_of_Cyan]

So I haven't given up on this problem just yet.

gneill your equations look very different from mine.

So for this:

Vx = 3Ω*I3 - 4V

would mean

and Vx - 3Ω*I3 + 4V = 0

but is the 4V positive for the reason you said that the resistor is shared with another mesh loop?

and don't see where your second equation comes from.

 

[gneill]

So I haven't given up on this problem just yet.

gneill your equations look very different from mine.

So for this:

Vx = 3Ω*I3 - 4V

would mean

and Vx - 3Ω*I3 + 4V = 0

but is the 4V positive for the reason you said that the resistor is shared with another mesh loop?

Yes, and the current in that other mesh flows counter the current in this mesh. So it generates a potential rise across the resistor as viewed from this mesh, as opposed to the potential drop caused by this mesh's current.

and don't see where your second equation comes from.

It corresponds to your mesh 2. Redo the "KVL walk" around the loop and you should arrive at the same thing. Remember to include the effect of current I1.

 

[Color_of_Cyan]

so that equation would mean:

Vx = (Vb - Va)Then Vx + 1Ω(i1 - i3) - 2Ω*i3 = 0

Vx + 1Ω(4A - i3) - 2Ω*i3 = 0

But why is it (i1 - i3) and not (i3 - i1)? Or is that just for consistency and it would also have to be (i1 - i3) for mesh 1 too?I think my only real problem with these is how to set up these equations properly, and even then it still gets hard solving them.

 

[gneill]

so that equation would mean:

Vx = (Vb - Va)


Then Vx + 1Ω(i1 - i3) - 2Ω*i3 = 0

Vx + 1Ω(4A - i3) - 2Ω*i3 = 0

But why is it (i1 - i3) and not (i3 - i1)? Or is that just for consistency and it would also have to be (i1 - i3) for mesh 1 too?

You just did mesh 3, not mesh 2. Current I3 runs in your mesh 3. Mesh 2 is the bottom left loop according to your circuit diagram in post #1.

When you "walk" around the loop you follow in the direction of your assumed mesh current. When you encounter a resistor there is a potential DROP in the direction of the current flow. If there is another mesh's current flowing in the same resistor then its current will be flowing in the opposite direction, so it produces a potential RISE in that resistor as you "walk" through it.

I think my only real problem with these is how to set up these equations properly, and even then it still gets hard solving them.

That comes with practice. Just be consistent about accounting for the drops in the direction of the mesh current, and rises for currents going against the direction of your "walk".

 

[The Electrician]

If you're doing your KVL walk around mesh 1, you pass through the rightmost 1Ω resistor from the right end of the resistor toward the left end.

If you're doing your KVL walk around mesh 3, you pass through the rightmost 1Ω resistor from the left end of the resistor toward the right end.

There's the answer to your question "But why is it (i1 - i3) and not (i3 - i1)?".

 

[Color_of_Cyan]

Okay. Do I still need 3 mesh equations or just one for mesh 2 now? I'm thinking the equation for that (going along with Vx and mesh 3 above) to be:

10V + (i1 - i2) - Vx = 0

10V + 1Ω(4A - i2) - Vx = 0(i1 - i2) going with what was said above. Would this be correct? Would I still need another equation?

I know there was also something about having to also do a supermesh instead if there's current source shared by 2 meshes; it wouldn't apply here?

 

[The Electrician]

If you're going to use an unknown for the voltage (Vx) across the 5 amp source, then you need 4 equations altogether.

If you use the supermesh technique, then you only need 3 equations, but for now why don't you stick with the Vx method.

At this point I think it would be good for you to list all your equations like this:

1.) I1 = 4
2.) I2 - I3 = 5
3.) ? (Mesh 2 equation)
4.) ? (Mesh 3 equation)

 

[Color_of_Cyan]

4.) I think would be:

10V + 4V - 1Ω*I2 - Vx = 0

= 14V - 1Ω*I2 - Vx = 0Didn't think i1= 4A would actually count as an equation, but okay, heh:1.) I1 = 4
2.) I2 - I3 = 5
3.) Vx - 3Ω*I3 + 4V = 0
4.) 10V + 1Ω(4A - i2) - Vx = 0

simplified:

4.) 14V - 1Ω*I2 - Vx = 0I already tried combining equations 3 and 4 but can't isolate I2 and I3 to get 5, so I substitute instead:

Vx = 3Ω*I3 - 4V
Vx = 14V - 1Ω*I2

14V - 1Ω*I2 = 3Ω*I3 - 4V

18V = 3Ω*I3 + 1Ω*I2I2 = 5A + I3

18V = 3Ω*I3 + 1Ω(5A + I3)

18V = 3Ω*I3 + 5V + 1Ω*I3

23V = 4Ω*I3

I3 = (23/4)A but still wrong. I think it might be an error with eq 4.

 

[The Electrician]

4.) I think would be:

10V + 4V - 1Ω*I2 - Vx = 0

= 14V - 1Ω*I2 - Vx = 0


Didn't think i1= 4A would actually count as an equation, but okay, heh:


1.) I1 = 4
2.) I2 - I3 = 5
3.) Vx - 3Ω*I3 + 4V = 0
4.) 10V + 1Ω(4A - i2) - Vx = 0

simplified:

4.) 14V - 1Ω*I2 - Vx = 0


I already tried combining equations 3 and 4 but can't isolate I2 and I3 to get 5, so I substitute instead:

Vx = 3Ω*I3 - 4V
Vx = 14V - 1Ω*I2

14V - 1Ω*I2 = 3Ω*I3 - 4V

18V = 3Ω*I3 + 1Ω*I2


I2 = 5A + I3

18V = 3Ω*I3 + 1Ω(5A + I3)

18V = 3Ω*I3 + 5V + 1Ω*I3

23V = 4Ω*I3

I3 = (23/4)A but still wrong. I think it might be an error with eq 4.

When you moved the +5V over the left side of the equation, you added 18 and 5, but you should have subtracted 5 from 18.

Your result should be:

13V = 4Ω*I3

I3 = (13/4)A

How does that check out?

 

[Color_of_Cyan]

It's right now. Thanks (and my bad).