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Please help - Stupid question about mesh-current analysis

There must be a very simple stupid answer to this and I could really
use some help knowing what it is :/

Consider a mesh-current analysis problem where the circuit contains a
current source. Every text and website I can find on the topic gives a
nice simple example where the current source is along the junction
between two meshes, and I understand how to set up a phantom voltage
across this source, which is canceled out by adding the two equations
for the meshes that adjoin.

However I do not for the life of me understand how to set up the
equations for such a circuit where the current source is on the edge
of the circuit. It's driving me crazy!

I've put an example of the sort of problem I mean at <http://
www.larwe.com/ebay/mesh.jpg>. I think (THINK) I can correctly deal
with the 10A source on the top edge, but the dependent source on the
bottom right corner has me utterly baffled. I assume KCL is used to
determine that the current through the 1 ohm resistor is (2Vphi/5) +
10 but what's the current in the 5 ohm?

Or in general what is the plan of attack when faced with a current
source at the periphery of a problem where mesh-current analysis has
been mandated?

(This detestable text has no solutions for the "drill exercises", just
answers. Useless).

Thanks for any help you can provide...
 
T

Tom Biasi

Jan 1, 1970
0
There must be a very simple stupid answer to this and I could really
use some help knowing what it is :/

Consider a mesh-current analysis problem where the circuit contains a
current source. Every text and website I can find on the topic gives a
nice simple example where the current source is along the junction
between two meshes, and I understand how to set up a phantom voltage
across this source, which is canceled out by adding the two equations
for the meshes that adjoin.

However I do not for the life of me understand how to set up the
equations for such a circuit where the current source is on the edge
of the circuit. It's driving me crazy!

I've put an example of the sort of problem I mean at <http://
www.larwe.com/ebay/mesh.jpg>. I think (THINK) I can correctly deal
with the 10A source on the top edge, but the dependent source on the
bottom right corner has me utterly baffled. I assume KCL is used to
determine that the current through the 1 ohm resistor is (2Vphi/5) +
10 but what's the current in the 5 ohm?

Or in general what is the plan of attack when faced with a current
source at the periphery of a problem where mesh-current analysis has
been mandated?

(This detestable text has no solutions for the "drill exercises", just
answers. Useless).

Thanks for any help you can provide...

Why should you care where the current source is?
You have three loops, draw the current in all three loops (take care with
your direction), and write you equations.
Post them here.
Tom
 
Why should you care where the current source is?

Because the algorithm given in the book for dealing with a current
source in this type of problem can't be applied to this configuration
- how do I draw a supermesh around something that's on the outside?
You have three loops, draw the current in all three loops (take care with

I'm really confused. This again doesn't follow the algorithm in the
text. if we define Ia as shown in the diagram, Ib as being
counterclockwise in the top loop, and Ic as being clockwise in the
right-hand loop, then

Ia = ???
Ib = 10
Ic = 2Vphi/5

According to the text I should be setting up voltage equations using
the mesh currents, and I can only draw this up for the bottom left
loop:

75 = 2(Ia+10) + 5(Ia - 2Vphi/5)

Both other loops have current sources in them, which I can't eliminate
by drawing a supermesh around them.

These problems have me totally baffled.
 
Let ib be the current in the right-hand mesh, which is also
coincidentally the current in the controlled source.  Then:

V_phi = (5)(ib - ia)

I think I just went further down the rabbit hole. Ia is flowing
clockwise. Should Vphi therefore not be 5(Ia-Ib)?

I have been up for 38 hours without sleep so far.. maybe this is
something to do with it.
 
Possibly.  I always get my signs backward at least once before I find and
fix the problem.
Go sleep!

Can't sleep until I work this out. This whole class of problems has me
pissed off, not the least reason for which being the textbook:

- it has answers only for some problems (usually these days textbooks
give answers for all odd problems, but this book has answers for only
maybe 30% of the questions)
- worked examples are given only for the simplest possible case of
each problem
- "drill exercises" are often more complex, but have only answers, no
intermediate values or working. STUPIDITY.

If they want to sell solution manuals, f#$*#!ing well put the solution
manual inside the textbook and add $30 to its price. Students don't
have a choice anyway.

Hate hate hate.
 
There must be a very simple stupid answer to this and I could really
use some help knowing what it is :/

I still hate this textbook, but I had a glass of shiraz and attacked
the problem again. Define Ib=10A running counterclockwise in the top
mesh, Ic=2Vphi/5 running clockwise in the lower right mesh and Ia
running clockwise in the lower right and we get:

75=2(Ia+10) +5(Ia-2Vphi/5)
so 7Ia - 2Vphi = 55

But Vphi=5(Ia-2Vphi/5)
so Vphi=5Ia/3

hence 7Ia - 10Ia/3 = 55
Ia = 15

*sigh* not enough hours in the day, I tell you.
 
T

Tom Biasi

Jan 1, 1970
0
There must be a very simple stupid answer to this and I could really
use some help knowing what it is :/

I still hate this textbook, but I had a glass of shiraz and attacked
the problem again. Define Ib=10A running counterclockwise in the top
mesh, Ic=2Vphi/5 running clockwise in the lower right mesh and Ia
running clockwise in the lower right and we get:

75=2(Ia+10) +5(Ia-2Vphi/5)
so 7Ia - 2Vphi = 55

But Vphi=5(Ia-2Vphi/5)
so Vphi=5Ia/3

hence 7Ia - 10Ia/3 = 55
Ia = 15

*sigh* not enough hours in the day, I tell you.

It would serve you to always represent your units.
75 is 75 v, 2 is 2 ohms. Not nit picking but it will help you realize what
units you are working with.
BTW: Ia is in the lower left. Good night.
Tom
 
J

Jon Kirwan

Jan 1, 1970
0
I still hate this textbook, but I had a glass of shiraz and attacked
the problem again. Define Ib=10A running counterclockwise in the top
mesh, Ic=2Vphi/5 running clockwise in the lower right mesh and Ia
running clockwise in the lower right and we get:

75=2(Ia+10) +5(Ia-2Vphi/5)
so 7Ia - 2Vphi = 55

But Vphi=5(Ia-2Vphi/5)
so Vphi=5Ia/3

hence 7Ia - 10Ia/3 = 55
Ia = 15

*sigh* not enough hours in the day, I tell you.

I don't really like mesh analysis much. There are several methods,
but the one I like conceptually uses the idea of "spilling" in and out
of a node.

In your schematic example, you have four nodes. The schematic is:
: ,------( <-- )-------,
: | I=10A |
: | |
: | Rx Vy Rz |
:Vx +---/\/\--+---/\/\---+ Vz
: | 2 | 1 |
: | | ^
: --- \ / \ I=
: - V1 / Ry | 2/5 * Vy
: --- 75V \ 5 v
: - /
: | | \ /
: | | v
: | | |
: gnd gnd gnd

Okay, I admit it. It's not the best ASCII schematic around. Oh,
well.

Node Vx is obvious. It is: Vx = 75V and we're done.

Node Vy spills outward through three resistors. Treat the resistors
as conductances for this purpose. Also, the same thing happens in
reverse (super-position), so that Vx, Vz, and gnd spill back into Vy.
The spill in and spill out must be equal:

Spill Out Spill In
----------------- -------------
Vy: Vy*(Gx + Gy + Gz) = Vx*Gx + Vz*Gz
Vz: Vz*Gz + 2/5*Vy + 10A = Vy*Gz

That's about it.

Solving the Vz equation for Vz, we get:

Vz = (Vy*(Gz - 2/5) - 10A)/Gz

Substituting that into Vy equation and solving for Vy:

Vy = (Vx*Gx - 10A)/(Gx + Gy + 2/5)

The values of Gx=1/Rx, Gy=1/Ry, and Gz=1/Rz should be clear. Plugging
in the values, that solves out to Vy=25V and using that with the
solved Vz equation gives Vz=5V.

Which pretty much finishes it. I(Ry)=25/5=5A; I(Rx)=(75-25)/2=25A;
I(Rz)=(25-5)/1=20A. Your B source of 2/5*Vy is clearly 10A and that
leaves 15A for the 75V source. Labeled below:
: ,------( <-- )-------,
: | I=10A |
: | |
: | 25A-> 20A-> |
: | Rx Vy Rz |
:Vx +---/\/\--+---/\/\---+ Vz
: | 2 | 1 |
: | | ^
: --- \ / \ I=
: - V1 / Ry | 2/5 * Vy
: --- 75V \ 5 v
: - / = 10A
: | ^ | 5A \ / |
: | | | | v v
: | 15A | v |
: gnd gnd gnd

I like that way of looking -- treat each node as spilling outward and
inward through conductances, in or out via current source/sinks, and
let the chips fall where they do. I don't like mesh as it doesn't
sing in my mind like this way does.

Jon
 
Unfortunately he's working through a textbook section on mesh analysis.
(I usually prefer nodal analysis, too.  I know there's been times when

I found the node-voltage method much simpler to wrap my head around,
but I more frequently make mistakes with signs in this method :)
mesh has made more sense, but not often & I can't remember when it does!)

The text really isn't very helpful on the decision process either.
They say basically pick whichever method results in fewer simultaneous
equations. But since I'm doing them all on a calculator anyway, it
makes no difference to me whether there are 2, 3, 5 or 10 unknowns.
Though sometimes if there are lots of voltage or current sources,
choosing one or the other can mean less work.

It's all very moot since the whole reason I am struggling with this
stuff is because I don't do it manually. It's the work of a few
moments to set up a circuit in PSpice and run two simulations to get a
Thevenin equivalent or whatever I need. Come to think of it, there's
probably some macro feature to do that automatically.
 
J

Jon Kirwan

Jan 1, 1970
0
Unfortunately he's working through a textbook section on mesh analysis.

I'm self-taught, as this is just a hobby. By the time I was exposed
to mesh analysis, I'd already learned to handle things using the nodal
approach. I'd read "The SPICE Book" and that's how they illustrated
the way spice looks at things. It made a whole lot of sense at the
time. When I finally was exposed to the mesh approach because my son
was taking an electronics class, I just kept going "Ewwww! No way!
Why would anyone do this to themselves?" I could work through it,
forced to it, but I didn't like it. What I'd already learned was so
much easier to apply, both conceptually and in terms of manual labor
as well.
(I usually prefer nodal analysis, too. I know there's been times when
mesh has made more sense, but not often & I can't remember when it does!)

Well, I'd actually like to hear an example if one ever does come to
mind. I worked through quite a few problems with my son using mesh
and was never exposed to a single case where I liked it better.

Jon
 
J

Jon Kirwan

Jan 1, 1970
0
I found the node-voltage method much simpler to wrap my head around,
but I more frequently make mistakes with signs in this method :)

The nodal analysis I was exposed to is just a little bit different
than what The SPICE Book exposed me to, but close. They don't usually
express it as "spills" in and out, when talking about it, but instead
take a slightly less intuitive and slightly more rigorous approach in
their wording. It's just enough different from how SPICE appears to
approach things and how nodal analysis does that I consider them very
close cousins but not identical twins, so to speak.

But I don't ever recall making such mistakes with nodal analysis/the
spice approach. I think I actually have MORE problems with signs in
other cases (though I usually get them right, I think -- I'm not sure,
anymore, because the nodal approach just works and works and works,
when I need it and I don't seem to have errors with it so I just
haven't gone back that often to see if I'd still have sign problems in
mesh or branch analysis ... neither of which I developed much of a
stomach for.)
The text really isn't very helpful on the decision process either.

It's easy -- I think branch analysis and mesh analysis are always
harder on every problem done by hand. Unless you just love to play
with matrices, Cramer's rule, and determinants. Carefully done, mesh
analysis appears to play right into a nice matrix setup, so if you
have matrix math handy on your calculator it can be a little less
effort arranging terms to plug in. But if you are working by hand,
paper and pencil, I don't like it at all. It's just _more_ work, not
less.

I now remember a conversation with Dr. Fouts, who's famous for
chimpanzee communications (he's well into his 70's now, but probably
still at Central Washington University -- I make contributions to his
institute there.) We were talking about some of the other approaches
taken by other researchers -- the use of plastic chips of various
colors or shapes, for example, to facilitate communication between
humans and chimps in lieu of using hand-signs. Dr. Fouts was almost
livid... angry... in response. He told me that it was cruel to the
animals to do that, because, "What happens when they don't have their
plastic chips around? How do they communicate?" He added that the
chimps get frustrated and even angry when that happens. He felt it
was cruel to teach an animal such boons in communication and force
them to develop skills with tools they cannot always have around --
their hands.

We don't always have matrix math calculators at hand. But we usually
do have paper and pencil, and if not even that we've got our fingers
and dirt to play with. The nodal analysis approach won't leave you
frustrated from a lack of tools, nor does it force you quite as often
towards matrix math and Cramer's. And it "feels" right, too.
They say basically pick whichever method results in fewer simultaneous
equations.

I'm still waiting to see the case where mesh results in fewer. hehe.
But since I'm doing them all on a calculator anyway, it
makes no difference to me whether there are 2, 3, 5 or 10 unknowns.
Though sometimes if there are lots of voltage or current sources,
choosing one or the other can mean less work.

Read above.
It's all very moot since the whole reason I am struggling with this
stuff is because I don't do it manually. It's the work of a few
moments to set up a circuit in PSpice and run two simulations to get a
Thevenin equivalent or whatever I need. Come to think of it, there's
probably some macro feature to do that automatically.

Well, spice is a great tool. When you have it handy. There will be
those times when it's not around.

Jon
 
The nodal analysis I was exposed to is just a little bit different
than what The SPICE Book exposed me to, but close.  They don't usually

I don't know if it's all schools or just this one, but the course is
really presented as "this is what's happening inside your SPICE
program". Not explicitly, but the implication is definitely there.
But I don't ever recall making such mistakes with nodal analysis/the
spice approach.  I think I actually have MORE problems with signs in

I make a lot of silly algebraic mistakes of this sort. The reason I
get signs wrong in the node method is because you have a lot of
situations like this:

V1---r1---V2---r2---V3
|
r3
|
V4

Let's say we define current flow to be V1->V2, V2->V3 and V2->V4.
Well, the very first term of the expression will be (V1-V2)/r1, but
because it's entering the node it will be negative, and the obvious
step is to rewrite -(V1-V2)/r1 as (V2-V1)/r1. I often make mistakes
there.

It's easy -- I think branch analysis and mesh analysis are always
harder on every problem done by hand.  Unless you just love to play
with matrices, Cramer's rule, and determinants.  Carefully done, mesh

OMG Cramer's rule... I learned it nearly 20 years ago, and never used
it, and was re-exposed to it in this course, but could not apply it
with a gun to my head.

animals to do that, because, "What happens when they don't have their
plastic chips around?  How do they communicate?"  He added that the
chimps get frustrated and even angry when that happens.  He felt it

I think the difference between his situation and mine is that if I'm
reduced to scrawling algebra in the dust, I won't have the physical
tools to do anything with the results anyway. I guess after the alien
abduction and anal probing, maybe I can impress my captors by doing
circuit analysis in the steam on the operating theater windows?
Well, spice is a great tool.  When you have it handy.  There will be
those times when it's not around.

True...ish. Rationally, I really only use this information in the
design phase, and since design no longer means Bishop Graphics I'll
always have CAD tools available when I need them.
 
J

Jon Kirwan

Jan 1, 1970
0
I don't know if it's all schools or just this one, but the course is
really presented as "this is what's happening inside your SPICE
program". Not explicitly, but the implication is definitely there.

Well, "The SPICE Book" is recommended by Mike Englehardt who is the
long time maintainer/programmer/electronics guru of LTSpice... and
he's one to know, I think. And right on page 15, an example is worked
that uses the nodal approach.

My guess here, if I had to make one, is that mesh analysis, because of
how the equations lay out neatly looking like they are matrix-ready at
the outset, that a lot of people just assume that is how spice
approaches things. They may be wrong about that guess.

But I'm not the expert here. What I do know about spice is that it
doesn't only use one tool. It must "linearize" around operating
points, move the dots for a tiny dt, re-linearize again around the new
operating point, etc. Deep inside, I suspect spice programmers use
all the tools they can muster including ways of recognizing when one
approach or another is more likely to yield better accuracy per unit
time. So maybe spice uses more than one approach. But if only one,
then from The SPICE Book I take the idea that they approach it
nodally.
I make a lot of silly algebraic mistakes of this sort. The reason I
get signs wrong in the node method is because you have a lot of
situations like this:

V1---r1---V2---r2---V3
|
r3
|
V4

Let's say we define current flow to be V1->V2, V2->V3 and V2->V4.
Well, the very first term of the expression will be (V1-V2)/r1, but
because it's entering the node it will be negative, and the obvious
step is to rewrite -(V1-V2)/r1 as (V2-V1)/r1. I often make mistakes
there.

I don't even think that way, at all!! Make NO assumptions.

V2 spills away through r1, r2, and r3. V1 spills in through r1. V2
spills in through r2. V4 spills in through r3. That's all there is.
No signs, yet. Just perspective. Putting it into a concrete equation
is trivial:

V2*(G1+G2+G3) = V1*G1+V4*G3+V3*G2

Done. It reads: "V2 times the sum of the conductances away from it
equals V1 times the conductance inward plus V4 times the conductance
inward plus V3 times the conductance inward." What's hard? And
notice that I didn't start by thinking in any particular direction. On
the contrary, I decided that everything flows in ALL directions at
once. They flow IN and OUT at the same time so it's not possible to
mess up on signs. Well. I think so.

Solving for v2, if you know the rest:
V2 = (V1*G1+V4*G3+V3*G2)/(G1+G2+G3)

What's even neater is that this reads nice, as well. "The voltage at
V2 is equal to the sum of the surrounding voltages times their
conductances divided (which is amps) times the equivalent parallel
resistance of those conductances (which is what 1/[G1+G2+G3] is.)"

It's just nice stuff everywhere you look.

Mesh doesn't form beuatiful pictures in your mind. Nodal does.
OMG Cramer's rule... I learned it nearly 20 years ago, and never used
it, and was re-exposed to it in this course, but could not apply it
with a gun to my head.

Hehe. Yeah. I had to refresh my memory, too. It's handy, though, if
you don't want the complete solution and just need a selected value or
two.
I think the difference between his situation and mine is that if I'm
reduced to scrawling algebra in the dust, I won't have the physical
tools to do anything with the results anyway. I guess after the alien
abduction and anal probing, maybe I can impress my captors by doing
circuit analysis in the steam on the operating theater windows?

;)

Well, I guess I find myself without a calculator or computer while
doing design more often than you. I don't carry laptops, TI-92's,
etc., to dinner at a restaurant... but if I'm stuck waiting in the
lobby for other guests to arrive first, I can sit there with pen and
paper and do stuff without feeling "locked out" by not having my
favorite calculator or computer around. Other times, I'm being driven
somewhere (might be a few hours' trip) and if I've forgotten (more
common these days) or otherwise didn't "bring my tools" with me, I can
think and strategize about things and even work out some of the
numbers when I want to.
True...ish. Rationally, I really only use this information in the
design phase, and since design no longer means Bishop Graphics I'll
always have CAD tools available when I need them.

Okay. You think you will always have your plastic chips around when
you want them. I can go with that. I guess things have worked out
often enough with me that if I depended on them exclusively I'd be
frustrated a lot more than I actually am. It's nice to need only your
brain and a few marks in the sand, at times.

Jon
 
J

Jon Kirwan

Jan 1, 1970
0
<snip>
What's even neater is that this reads nice, as well. "The voltage at
V2 is equal to the sum of the surrounding voltages times their
conductances divided (which is amps) times the equivalent parallel
resistance of those conductances (which is what 1/[G1+G2+G3] is.)"
<snip>

Okay, yeah. Bad sentence. I meant, "The voltage at V2 is equal to
the sum of the surrounding voltages times their conductances (which is
amps) times the equivalent parallel resistance of those conductances
(which is what 1/[G1+G2+G3] is.)" That _divided_ was my lower
functioning brain reading '/' and telling my fingers to do something
faster than my higher functioning part could read 1/conductance and
see ohms and then tell my fingers to write _times_... Weird thing is,
both got their way with me. ;)

Jon
 
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