Two black boxes display the same impedance at all frequencies. Which has the single resistor?

Question

Two black boxes display the same impedance at all frequencies. The first contains a single 1 Ohm resistor. Each end is connected to a wire, so that two wires protrude from the box. The second box looks identical from the outside, but inside there are 4 components. A 1 F capacitor is in parallel with a 1 Ohm resistor, and a 1 H inductor is in parallel with the other 1 Ohm resistor. The RC combo is in series with the RL combo, as shown in the figure

The boxes are painted black, unbreakable, impervious to x-rays, and magnetically shielded.
Demonstrate that the impedance of each box is 1 Ohm at all frequencies. What measurement would allow one to determine which box contains the single resistor?

Answer 1

This is an addendum to luchador's answer.

The transient power dissipation in the two boxes is very different. The following simulation demonstrates this.

^{simulate this circuit – Schematic created using CircuitLab}

Run the simulation for 40 seconds, and plot the expression "I(R1.nA)^2+I(R2.nA)^2", which represents the total instantaneous power in the two resistors.

As I said in my comment, box A will not only heat up more slowly while the pulse is on, it will exhibit a spike in temperature when the pulse ends, because the total instantaneous power dissipated in the resistors is doubled at that moment. Box B will not exhibit such a spike.

(NOTE: If you have trouble running the simulation, see this Meta post.)

Answer 2

The only observable difference is lagged dissipation of power as heat. Any restriction of observing heat transfer is against thermodynamics laws. So, somehow you can observe that and figure out, despite of that restrictions list.

Answer 3

Measure the thermal noise of the resister and you will get KTB from college or damn close to it. The box with the reactive components will give some measurable noise too BUT it is the vector sum of HF rolled off and LF rolled off noise. The math is a bit long for this but suffice to say that there will be a difference in your noise measurements. On a spectrum analyser you will see some lack of flatness around the resonant frequency. Because the network has a Q of 1 the effect will be quite broad. If you wanted to do this as an actual experiment and not just an experiment in thought you will need to choose component values that would be more physically realisable and easier to make more ideal .

Answer 4

You could apply a DC voltage to box A. That will charge the capacitor. Now you can remove the source and measure the stored voltage. That does not work for box B.

Update: For this particular choice of components the system is not observable. For this reason this method won't work. When we apply a voltage to the circuit, we'll have a current through the inductor and a charge on the capacitor. As soon as we remove the voltage, the inductor's current will flow through the parallel resistor, thus cancelling the voltage on the capacitor. The current of the inductor and the voltage on the capacitor will decay at the same rate. They can't be observed from the outside.

Answer 5

In box A, \$R_L\$ is in parallel with \$L\$, which has some DC resistance, \$ R_{(L)}\$.

The total resistance of \$R_L\$ and \$ R_{(L)}\$, then, is:

$$ Rt = \frac{R_L \times R_{(L)} }{R_L \ + R_{(L)}} \text { ohms,} $$

which must be less than \$R_L \Omega\$ but greater than \$ 0\Omega \$.

\$R_T\$ is in series with \$R_C\$, so their total resistance must be greater than one ohm.

Box B, however contains a one ohm resistor, so the identities of the boxes can be confirmed by measuring the end-to-end resistances of the wires protruding from the boxes, with box A exhibiting a higher resistance than box B.

Answer 6

Make a third terminal by tightly enclosing the current box with a metal box (or just use the current box if it is already metal). Then, measure the frequency response of each of the original two terminals with respect to this new terminal: Box B's responses should be more symmetric (Box A should show some difference depending upon whether you probe the capacitor terminal or the inductor terminal).

I doubt you can design two boxes such that they are indistinguishable for this three-terminal experiment. Please give box details if you can.

Answer 7

Let's assume to start with that the components are well enough matched, which itself is an issue given tolerances on capacitors and inductors.

You're assuming an ideal inductor. In the real world, the inductor core goes into saturation with enough current/frequency applied. Unless you have an air-core inductor, of course, but that will always radiate in various interesting ways which are detectable externally.

You're also assuming the capacitor is not polarised and has no breakdown voltage. Polarisation is easy to check - simply put a negative voltage across it. Breakdown voltage may be harder, given that we would need a lot of current as well. The obvious solution there though is that a step-change in current (a hard switch-off) will produce a massive voltage spike from the inductor. That's how a car's spark plugs are driven, producing several kV from a 12V battery. Doing the same here would likely push the capacitor beyond its breakdown voltage.

Answer 8

Hook up a time-domain reflectometer and send a pulse into the box. The reflections should show the presence of multiple elements.