Nodal analysis problem

mo2015mo · May 9, 2013

Hi guys,

I have been having problems with the attached question , I have tried to solve it and write this 3-nodal equations as below

node 1: 5 V1 + 0.2 V3 + 2(V1 -V2) =0
node 2: -0.2V3 + 10 V3 + 20(V2-V3) =0
node 3 : 20 (V3-V2) + 2(V3-V1) -3 V2=0

-there are dependent sources in question But i don't find the relation between it.

Could you guide me about my mistake?

The Electrician · May 9, 2013

mo2015mo said:
Hi guys,

I have been having problems with the attached question , I have tried to solve it and write this 3-nodal equations as below

node 1: 5 V1 + 0.2 V3 + 2(V1 -V3) =0
node 2: -0.2V3 + 10 V2 + 20(V2-V3) =0
node 3 : 20 (V3-V2) + 2(V3-V1) -3 V2=0

-there are dependent sources in question But i don't find the relation between it.

Could you guide me about my mistake?

I see a couple of errors (shown in red) and the terms in blue don't belong where you have them because they aren't currents.

mo2015mo · May 9, 2013

ok thanx ALOT ALOT 4 u,,

But when find the input admittance (Yin) between node 1 and the reference as stated in the question,
my answer does not identical with Book answer

plz trace my answer method,,

node 1: 5 V1 + 0.2 V3 + 2(V1 -V3) =0 ==> 7 V1 - 0.8V3 =0
node 2: -0.2V3 + 10 V2 + 20(V2-V3) =0 ==> 30 V2 - 20.2 V3=0
node 3 : 20 (V3-V2) + 2(V3-V1) -3 V2=0 ==> -2V1 - 23 V2 +22V3 =0

so, determinant ΔY = 1319.8 S^3
and Δ11 = ΔY11 = 195.4 S^2

thus, Yin = ΔY / Δ11
= 1319.8/195.4 = 6.75 (In my book the right answer equal 10.68 )

Could you guide me about my mistake?

WBahn · May 9, 2013

mo2015mo said:
Hi guys,

I have been having problems with the attached question , I have tried to solve it and write this 3-nodal equations as below

node 1: 5 V1 + 0.2 V3 + 2(V1 -V2) =0
node 2: -0.2V3 + 10 V3 + 20(V2-V3) =0
node 3 : 20 (V3-V2) + 2(V3-V1) -3 V2=0

Always, always, ALWAYS track your units!

The problem is a bit sloppy about the units, which is a shame and, unfortunately, not uncommon. But you can path that up before you even begin.

The dependent source on the left is a voltage source and the controlling equation is 0.2V3, so this one is fine. However, the dependent source on the right is a current source and has a controlling equation given as 3V2. Well, this is nonsense because this has units of voltage and the source is producing a current. You have to make an assumption here that the units of the coefficient are actuall "amperes/volt".

The the admittances have units of seimens, which is also amperes/volt, hence putting the units into the equations as you have them you have:

node 1: (5A/V) V1 + 0.2 V3 + (2A/V)(V1 -V2) =0
node 2: -0.2V3 + (10A/V) V3 + (20A/V)(V2-V3) =0
node 3 : (20A/V) (V3-V2) + (2A/V)(V3-V1) -(3A/V) V2=0

Now, look at the units of each term. Notice that the terms that The Electrician highlighted in blue have units of voltage, while all the others have units of current. Since you can't add voltage and current, you KNOW that these equations are wrong. If you track your units, most (not all) of the mistakes you make will SCREAM out to you, "Hey, I'm wrong!" There's no point going any further until you track down and correct the error. Wouldn't you rather do it immediately after you make the error rather than after a few pages of algebraic gymnastics that have been a waste of time from the start?

Iin figuring out why those terms are wrong, you will be forced to confront what it is you are fundamentally trying to do, which is apply KCL at each node, and you'll figure out how to do it correctly.

Note that, because of the voltage source, the proper way to apply Nodal Analysis if to define a supernode surrounding this source.

WBahn · May 9, 2013

mo2015mo said:
ok thanx ALOT ALOT 4 u,,

But when find the input admittance (Yin) between node 1 and the reference as stated in the question,
my answer does not identical with Book answer

plz trace my answer method,,

node 1: 5 V1 + 0.2 V3 + 2(V1 -V3) =0 ==> 7 V1 - 0.8V3 =0
node 2: -0.2V3 + 10 V2 + 20(V2-V3) =0 ==> 30 V2 - 20.2 V3=0
node 3 : 20 (V3-V2) + 2(V3-V1) -3 V2=0 ==> -2V1 - 23 V2 +22V3 =0

Wait a minute. The Electrician specifically pointed out that you had terms that didn't belong or were wrong and you just ignored that and plugged away using equations that you were TOLD were incorrect and now want people to trace through them and tell you what's wrong? What's wrong is that your equations are wrong from the very get go!

mo2015mo · May 10, 2013

Yes ... yess
i will track the units ,thanx alot dr. WBahn

Really did not learn supernode way, only yesterday
i used supernode as below

V1 -V2 = 0.2V3

node 3 : 20 (V3-V2) + 2(V3-V1) -3 V2=0

supernode : 5 V1 + 10V3 ++2 (V1-V2) +20(V2-V3)=0

BUT there is the same problem faced me and it's the units??
V1 -V2 = 0.2V3 ==> in Ampere

and Now from 3 equations written above How can find determinant ΔY and Δ11 ??

i waste 3 hours of my time every day to solve this question last four days,and Did not get the correct answer.

and my final exam on next thursday.

plz Help meeeeeee

The Electrician · May 10, 2013

You are very close to the solution. You just have a couple of errors.

mo2015mo said:
Yes ... yess
i will track the units ,thanx alot dr. WBahn

Really did not learn supernode way, only yesterday
i used supernode as below

V1 -V2 = 0.2V3

node 3 : 20 (V3-V2) + 2(V3-V1) -3 V2=0

supernode : 5 V1 + 10V2 ++2 (V1-V3) +20(V2-V3)=0

BUT there is the same problem faced me and it's the units??
V1 -V2 = 0.2V3 ==> in Ampere

and Now from 3 equations written above How can find determinant ΔY and Δ11 ??

The units on your first equation (blue) are ok because it is not a KCL equation; it's just a constraint equation. The units for everything are volts. Where you had a units problem before, you were mixing volts and amps; here you just have volts.

You've made the same kind of error that I pointed out in post #2. You have V2 where you should have V3, and V3 where you should have V2. I've designated them in red. You better be extra careful with this on your final.

What you need to do now is collect terms and get your equations in the form of a matrix. Let's put the supernode equation second:

Constraint equation: V1 -V2 = 0.2V3 --> V1 - V2 - .2 V3 = 0

Supernode : 5 V1 + 10V2 + 2 (V1-V3) +20(V2-V3)=0 --> (5+2)V1 + (10+20)V2 - (20+2)V3 = 0

Node 3: 20 (V3-V2) + 2(V3-V1) -3 V2=0 --> -2V1 - (20+3)V2 + (2+20)V3 = 0

In matrix form this will be:

Rich (BB code):

[ 1   -1 -.2 ]
[ 7   30 -22 ]
[-2  -23  22 ]

Now, you might think that the admittance at node 1 will be ΔY/Y11, but that won't work because row 1 of your matrix is not a KCL equation, but rather just a constraint equation. Row 2 is the KCL equation for the supernode, so the admittance at node 1 is given by -ΔY/Y21 and the admittance at node 2 is ΔY/Y22, both expressions using cofactors from row 2.

Do you know how to use a linear solver, rather than cofactors, to solve a system of linear equations?

See if you can get the correct result now that you have a supernode equation. Let us know your result.

WBahn · May 10, 2013

mo2015mo said:
Yes ... yess
i will track the units ,thanx alot dr. WBahn

Really did not learn supernode way, only yesterday
i used supernode as below

V1 -V2 = 0.2V3

node 3 : 20 (V3-V2) + 2(V3-V1) -3 V2=0

supernode : 5 V1 + 10V3 ++2 (V1-V2) +20(V2-V3)=0

You say you will track units and then you immediately proceed to completely ignore units. Not a good way to make progress.

The admittances all have units of A/V. The gain of the dependent current source also has units of A/V because it is a transconductance amplifier.

BUT there is the same problem faced me and it's the units??
V1 -V2 = 0.2V3 ==> in Ampere

If you are paying attention to the units, you would know the answer. V1-V2 can only result in a voltage. The 0.2 multiplying the V3 is the gain of a voltage amplifier, hence it is dimensionless. Thus, the result of either equation is a voltage, not an amperage.

So let's look at how you SHOULD have written your equations:

V1 -V2 = 0.2V3

node 3 : (20A/V) (V3-V2) + (2A/V)(V3-V1) -(3A/V) V2=0

supernode : (5A/V) V1 + (10A/V)V3 +(2A/V) (V1-V2) +(20A/V)(V2-V3)=0

Now go through term by term and see if each term means what you wanted it to mean.

In particular, look at the second term of the supernode equation. That says that you have a current flowing out of the supernode that is equal to the voltage on node 3 multiplied by a 10S conductance. Does that make sense?

What about the third term. That is for the current flowing between V1 and V2 through a 2S conductance. Can you point to where you have that anywhere in your circuit?

i waste 3 hours of my time every day to solve this question last four days,and Did not get the correct answer.
and my final exam on next thursday.
plz Help meeeeeee

You need to learn, really learn, the two cardinal rules of practical engineering:

1) Always, always, ALWAYS track your units.
2) Always, always, ALWAYS ask if the answer makes sense.

Neither of these apply just to the final result. But should be applied at every step of the way. Don't just write down an equation and press on. Review the equation carefully, as if you were the grader, and verify that it really does make sense and that the units work out. Otherwise, you will spend hours and hours solving equations that are guaranteed never to work out.

It's the classic tradeoff of pennywise and pound foolish. You can spend a little bit of time up front to keep from getting stuck, or you can spend hours trying to dig yourself out of a hole that was easily avoidable. But you have to make the conscious choice to avoid the holes and develop the necessary habits to make doing so second nature.

The Electrician · May 10, 2013

WBahn said:
Now go through term by term and see if each term means what you wanted it to mean.

In particular, look at the second term of the supernode equation. That says that you have a current flowing out of the supernode that is equal to the voltage on node 3 multiplied by a 10S conductance. Does that make sense?

What about the third term. That is for the current flowing between V1 and V2 through a 2S conductance. Can you point to where you have that anywhere in your circuit?

The OP made this kind of mistake early on, and he made it again. I pointed this out to him in post #7, so he should be doubly aware of it now.

mo2015mo · May 10, 2013

The Electrician said:
You are very close to the solution. You just have a couple of errors.

Hey, I'm wrong!

Do you know how to use a linear solver, rather than cofactors, to solve a system of linear equations?

See if you can get the correct result now that you have a supernode equation. Let us know your result.

i found ΔY= 394.2
and Δ21 = -48.4 , Yin = - 394.2/-48.4 = 8.14 (Not correct ans.)

The Electrician · May 10, 2013

Sorry, there's a typo in the matrix! You should have caught that!

Here's your matrix:

Rich (BB code):

[ 1   -1 -.2 ]
[ 7   30 -22 ]
[-2  -23  22 ]

The determinant of this is 284.2

Try your calculations now.

mo2015mo · May 10, 2013

WBahn said:
In particular, look at the second term of the supernode equation. That says that you have a current flowing out of the supernode that is equal to the voltage on node 3 multiplied by a 10S conductance. Does that make sense?

What about the third term. That is for the current flowing between V1 and V2 through a 2S conductance. Can you point to where you have that anywhere in your circuit?

You need to learn, really learn, the two cardinal rules of practical engineering:

1) Always, always, ALWAYS track your units.
2) Always, always, ALWAYS ask if the answer makes sense.

Neither of these apply just to the final result. But should be applied at every step of the way. Don't just write down an equation and press on. Review the equation carefully, as if you were the grader, and verify that it really does make sense and that the units work out. Otherwise, you will spend hours and hours solving equations that are guaranteed never to work out.

It's the classic tradeoff of pennywise and pound foolish. You can spend a little bit of time up front to keep from getting stuck, or you can spend hours trying to dig yourself out of a hole that was easily avoidable. But you have to make the conscious choice to avoid the holes and develop the necessary habits to make doing so second nature.

okition , thanx 4 your advices

mo2015mo · May 10, 2013

The Electrician said:
Sorry, there's a typo in the matrix! You should have caught that!

Here's your matrix:

Rich (BB code):

[ 1 -1 -.2 ] [ 7 30 -22 ] [-2 -23 22 ]

The determinant of this is 284.2

Try your calculations now.

really really really i was ...

I will not repeat these mistakes in the coming days

i calculations it and the result is corrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrect finally

The Electrician & WBahn
thanks you for help me

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Nodal analysis problem

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