Hi, The Electrician,
I used the equations of BlackSuede by set I3=0 then solve them.
I got the same result. Is this method really wrong?

No, it's not really wrong. I tend to encourage students to learn the most general methods of solution because they always work. Sometimes you can see a shortcut and if you're sure it will work, take a chance.

What I do when I want to verify a solution to a network is to solve it in at least two ways and compare the answers. That's why it's good to know the standard methods of solution.

In the case of this problem, it might have been required to find a general expression for I3, and it doesn't hurt to find it since it can be used to solve the particular situation here.

 

Ah, thanks!
My mistake is that I have never thought about the voltage across it. Each time I see that I = 0, I usually consider it open.

I was trying to think of a more familiar example in which similar reasoning might be applied.

The Wheatstone bridge at balanced condition comes to mind. The bridge detector at balanced condition carries no current but it is certainly not an open circuit.

 

Attached is my approach with the final few steps removed - in deference to the homework rules.

It's a stretch of credibility to expect a student to solve this "in a few lines".

To me few would be less than five. Subjective as that might be.

In general I agree, with the caveat that we don't know what what presented before this in the course that was intended to foreshadow approaches like this.

While "a couple lines" is an overexageration, it can clearly be done with a very limited number of lines and, more importantly, without solving mutliple simultaneous equations or performing polar-rectangular conversions all day. But the insight into how to do it is a bit tricky, though it does fall out directly from the conditions stated in the problem.

But either example problems exploiting similar tricks should have been presented (and maybe they were) or some hints on this particular problem should have been provided.

 

... more importantly, without solving mutliple simultaneous equations ...

Why is it important that it be done without solving multiple simultaneous equations? Solving multiple simultaneous equations is the essence of network theory. It's something the student should know well.

But either example problems exploiting similar tricks should have been presented (and maybe they were) or some hints on this particular problem should have been provided.

Or the student could learn methods that don't rely on a trick; methods that always work.

 

I've got no problem with learning general methods that always work, but there is also something to be said for not using a sledgehammer when a feather will do. Developing the ability to look for aspects of a given problem that can be exploited is a very valuable skill. If you instead just get out the chainsaw and blindly apply it to every problem you see, then you reduce yourself to little more than an automaton. Instead, fill your toolbox with a variety of tools, both elegant and brute, and then learn how to select the best tool for the job. But if the only tool in your box is a hammer, then everything looks suspiciously like a nail.

 

I've got no problem with learning general methods that always work, but there is also something to be said for not using a sledgehammer when a feather will do. Developing the ability to look for aspects of a given problem that can be exploited is a very valuable skill. If you instead just get out the chainsaw and blindly apply it to every problem you see, then you reduce yourself to little more than an automaton.

The comparison of a sledgehammer to a feather greatly exaggerates the comparison of standard nodal analysis to the elimination of one node in this particular circuit.

Instead, fill your toolbox with a variety of tools, both elegant and brute, and then learn how to select the best tool for the job. But if the only tool in your box is a hammer, then everything looks suspiciously like a nail.

Nodal and mesh analysis can't be blindly applied. If it were that easy, none of the people who ask for help here with those methods would be here.

They are not brute force tools. They are quite elegant. Ever since Feldtkeller began applying matrix methods to circuit analysis they have been the standard methods taught to beginners.

The standard methods of analysis are what these people are just now learning and practice is what leads to facility with the tools. The only tools in their box right now should be the standard methods, and they should learn to use them well. They should apply them to every problem they have to solve. If they happen to see a trick then they should use it in addition to the standard method to verify the correctness of the trick. In this way they build up a catalog of verified tricks. But they should use the standard methods often; there's nothing wrong with practice.

 

Shouldn't Lx equal 1/(2wC)? not 1/(2w^2C)?

 

Shouldn't Lx equal 1/(2wC)? not 1/(2w^2C)?

Why do you think that to be the case?

 

Shouldn't Lx equal 1/(2wC)? not 1/(2w^2C)?

Check the units.

Lx has to have units of henries, but this can be expressed in more basic units as:

V = 1/2 L di/dt
L = 2 V dt/di => volt-seconds/ampere

Similarly

Q = CV
dq/dt = i = C dv/dt
C = Q/V = i dt/dv => coulombs/volt = amp-seconds/volt

ω has units of 1/seconds

Hence, the relationship you are suggesting results in

Lx equal 1/(2wC)

volt-seconds/ampere equal 1/[(1/second)*(amp-seconds/volt) ]

volt-seconds/ampere equal volt/amp

seconds equal 1

Which does not work out. It says that you need something on the right hand side that has units of seconds (or divide the right hand side by something that has units of inverse-seconds).

The formula that you think is wrong does exactly this. Now, that doesn't mean that it is correct, but it is dimensionally consistent, which is a requirement. Yours, on the other hand, is dimensionally inconsistent and therefore you know it is not correct.

 

The formula that you think is wrong does exactly this. Now, that doesn't mean that it is correct, but it is dimensionally consistent, which is a requirement. Yours, on the other hand, is dimensionally inconsistent and therefore you know it is not correct.

Bravo!

I sent specialk over here from another forum. Two people asked about it over there, and two asked about it here. That's a total of 4 people asking for help with this problem. It must be a popular problem this year.