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We were given the following question in our circuit theory class:

Find the relation between \$R,L,C\$ such that the net reactance across terminals \$A\$ and \$B\$ is \$R\$.

schematic

simulate this circuit – Schematic created using CircuitLab

I followed the generic method (equivalent parallel reactance):

$$\frac{1}{R}=\frac{1}{R+j\omega L} + \frac{1}{R+\frac{1}{j\omega C}}$$ to get the required relation (after eliminating \$\omega\$).

Upto that is okay. But I'm confused about something:

Suppose the system is inside a black box, with only terminals \$A\$ and \$B\$ outside. In that case how can we possibly distinguish between a black box containing this circuit vs. a black box which contains only a resistance \$R\$?

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    \$\begingroup\$ Well, what is the difference between reactance and resistance? What tools would you have in a lab that you could use to explore that? How can you measure just the resistive component? \$\endgroup\$
    – John D
    Commented Mar 8, 2018 at 15:59
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    \$\begingroup\$ In which way did you eliminate \$\omega\$? \$\endgroup\$
    – Arsenal
    Commented Mar 8, 2018 at 15:59
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    \$\begingroup\$ @Arsenal asks a good question, do you really believe that you can eliminate ω for all frequencies? \$\endgroup\$
    – John D
    Commented Mar 8, 2018 at 16:06
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    \$\begingroup\$ @JohnD I don't think this qualifies directly as homework question as the question at hand came up during solving of an assignment and the answer to the assignment is not part of the question at all. \$\endgroup\$
    – Arsenal
    Commented Mar 8, 2018 at 16:12
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    \$\begingroup\$ The only way I can come up with to distinguish the two circuits is to apply a DC voltage to the terminals. This charges the cap if you wait a few time constants. Then immediately remove the DC voltage and measure the voltage at the terminals. If it's zero the circuit is purely resistive. If you see a voltage that decays with time there's a reactance present. \$\endgroup\$
    – John D
    Commented Mar 8, 2018 at 17:50

1 Answer 1

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Connect a resistor between A and the output of a signal generator; other lead of signal generator to B. Connect channel 1 of your scope to the signal generator output and channel 2 of the scope to A. Sync the scope to channel 1. Vary the frequency and observe phase shift, and amplitude of A with respect to channel 1. If this doesn't make sense, I suggest you go read about LC resonance and Q. We can provide further assistance after you report your findings!

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  • \$\begingroup\$ This doesn't work. We've already demonstrated that the reactances cancel out! \$\endgroup\$
    – Dave Tweed
    Commented Mar 8, 2018 at 18:42
  • \$\begingroup\$ @Dave Tweed True...at resonance. But as you approach either side of resonance, XC and XL should be distinguishable. At some point >> resonance, C and R should prevail to yield a pole as evidenced by the 45 deg. shift and .707 drop in amplitude. \$\endgroup\$
    – AlmostDone
    Commented Mar 8, 2018 at 18:49
  • \$\begingroup\$ Try it. There is no resonance! And besides, this was thoroughly hashed out a couple of years ago in the duplicate question linked at the top of this page. Even though the specific circuit is different, the same arguments apply. \$\endgroup\$
    – Dave Tweed
    Commented Mar 8, 2018 at 18:55
  • \$\begingroup\$ The answer is have equal time constants, e.g. L = 220 uH, R = 100 ohms and C = 22 nF. If you ac-sweep this network impedance, the phase is 0° and the real part is 100 ohms. Follow the steps I gave in the comment section (factor R in the impedance expression and equalize N and D) and you're there in a few seconds. \$\endgroup\$
    – Verbal Kint
    Commented Mar 8, 2018 at 19:37
  • \$\begingroup\$ @DaveTweed You are right, thanks for your comment. \$\endgroup\$
    – AlmostDone
    Commented Mar 9, 2018 at 18:40