I'm stuck with an exercise and would appreciate any hints.

I have to calculate the impedance between the points A and B:

enter image description here

So what I got is (starting from the right):

2L + (2R || R) + (3C || 3C)

but I don't know how to deal with this bridge-like 2C and C capacitors.

I don't expect a solution from you guys, just a hint how to do it.



Note: Russell has a nice solution, which is simpler than mine, but only works if the problem is made for it (i.e. a trick question), while my solution is more generally applicable. I guess it depends on how far you got in your circuit analysis class. If you have the feeling this kind of circuit is over your head the professor probably made it a trick question with a simple solution.

Starting from the left you have a triangle L-2L-C. You can transform this using a delta-star transformation. Write down the transformation formulas from delta to star and vice versa in big letters, you're going to need them. Applying those you get an impedance in series with the 2C on the left, and the other legs of the star will be in series with the 3R+3R and 2R+R, resp. If you redraw your schematic at this stage you'll see a new triangle or delta, this time with the 2C in the center of your schematic as one of the sides. Again apply the delta-star transformation. A few more steps and some parallel impedances and everything is just one long chain.

  • \$\begingroup\$ you just saved my day, we haven't covered delta-star transformation in our lecture. This will save me a lot of time in the exam. Big Thanks! \$\endgroup\$
    – sled
    Jul 31 '11 at 12:51
  • \$\begingroup\$ @sled - or you were asleep when they covered it:-). I wouldn't know how to solve this without that transformation. Success! \$\endgroup\$
    – stevenvh
    Jul 31 '11 at 13:13
  • \$\begingroup\$ @stevenvh have you ever used this type of calculation in real app? I haven't and I am curious if you ever did or know somebody who did. \$\endgroup\$
    – Frank
    Jul 31 '11 at 15:28
  • \$\begingroup\$ @Frank - frankly, no, but then I never needed to: a circuit like this seems purely academic to me. I can imagine that the equation can become very ugly if you don't simplify after each step. \$\endgroup\$
    – stevenvh
    Jul 31 '11 at 15:37
  • \$\begingroup\$ @Sled, the fact that it has not been taught may be a sign there is a different approach they are trying to give a challenging problem for. Make sure you can do all of the approaches taught in class proficiently. \$\endgroup\$
    – Kortuk
    Aug 1 '11 at 1:44

As StevenH says, Star-Delta is your GREAT friend. Learn to use it and spot where it can be used.

BUT note that this is a trick question. There is no guarantee that the one in an exam will be, so you can't depend on the trick saving you as it does in this case.

As I note at the end - the examiner is leading you to see his (her) trick by providing series and parallel component pairs that cry out for simplification. Fo0llow the hints and see where they lead.

What I'm going to do below blows right through the heart of this "problem". Consider this a "spoiler". I would not solve people's numerical homework for them, but this is somewhat different. You (Sled) decide - there is something about this problem which utterly dismantles its apparent complexity. In some ways that is probably the point of the exercise - can you spot that a problem can be reduced to something much simpler than is apparent? In real life this is not as clear cut as here BUT often of more importance - ie if you can look through apparent complexity to see the core problem underneath.

So - I'd suggest you stop reading this now, take the example and simplify it down as much as you can, then stare at it to see what the "trick" is, then come back here once you have seen it, or really can't see it (preferably the former).






  • First note that the 2R and R at bottom are in parallel and can be combined.

  • So too the next 3C and 3C

  • The 3R + 3R = 6R - combine them.

  • So too the 2R + R = 3R.

    Now, the fun begins.

  • Down the lower leg you have R + 3R + 3L with tappings.

  • Down the upper leg you have 2R + 6R + 6L = 2 x (R + 3R + 3L)

ie the upper leg and lower leg if treated separately have equal potentials at each of the intermediate nodes, so the C and 2C joining capacitors have no current and can be replaced by open circuits OR short circuits !!!. If the latter then you can combine elements in each parallel path and arrive at the single series string that StevenH's star-delta got to - but with no transformations and calculations (apart form simple parallel combinations.)

Again - you can't depend on them doing this in an exam of in real life :-)

How did I spot this?: I looked for it. Examiner's and test setters tend to do such arcane things for whatever reason. The devices in parallel with hard links are a signal to you that you can and need to simplify. So too the components in series. The examiner is both telling you something and hiding something. Accept his leads. Worst case you get a simpler diagram. Best case you can (almost) solve it in your head.

  • \$\begingroup\$ Well spotted! It may explain how the professor would ask put a problem like this before they learned about star-delta transformation. In my college life I was never so lucky to have this kind of trick questions that you could simplify this way. Also, one component with a different value, and the method is useless. \$\endgroup\$
    – stevenvh
    Aug 1 '11 at 9:22
  • \$\begingroup\$ One the one hand this is a good exercise to learn how to spot this kind of simplifications, but on the other hand, IRL you'll never have this kind of situation, so in the end the problem looks a bit pointless to me. What's the use of knowing how to solve trick questions if you can't solve the real ones? (A critique to the professor, not Russell) \$\endgroup\$ Aug 1 '11 at 10:26
  • \$\begingroup\$ hi, thanks that's probably the wanted solution ;) Actually I'm studying mechanical engineering and this exercise is from an old exam. The joke is that you can solve every exam in half the time if you can "see" the trick. So the best solution is to look for the trick and apply delta-star transformation when in doubt :) \$\endgroup\$
    – sled
    Aug 1 '11 at 10:50

One solution that many people forget depends on knowledge of other approaches. Many students have already mastered either nodal analysis(KCL) or mesh analysis(KVL).

If you need the impedance of a completely passive network you can apply a test voltage(call it 1V if it helps) and then solve for current using one of these methods. This is not necessarily the fastest method, but this is a method you already know.

When you complete V=IZ so Z=V/I. Problem solved, no fancy new tricks, but you used a basic method that works dependably.


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