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We are given a black box that contains only linear circuit elements and a pair of ports. We conduct the following two experiments with this black box.

  1. With the right port open, we applied V1 = 2V to the left port and measured I1 = 4mA and V2 = 8V.

  2. With the left port shorted, we applied I2 = -10 mA to the right port and measured I1 = 5mA and V2 = -10V.

Now we connect a current source to the right port and a voltage source to the left port. I1 = 8mA and V2 = 4V now. Find I2 and V1, in mA and V each.

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I don't get this. Since we have analyzed the circuit separately for each source, adding them should give me the final values of I1, V2, I2, V1 etc. However, that's not quite true—from cases 1 and 2, I should have I1 = 4 + 5 = 9 mA, but it's 8mA. Hence, I thought that the black box consumes 1mA; hence I2 = 0 + (-10) - 1 = -11 mA. Similarly, V2 = 4V even though adding the V2 from first two cases gives -2; hence, the black box provides an additional 6V, and so V1 = 2 + 0 + 6 = 8V. However, this approach is wrong.

How might I solve this?

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  • \$\begingroup\$ The applied voltage and current you need to find the results for are different values than were used for the tests, so you can't just add the test results together to get the final answer. \$\endgroup\$
    – The Photon
    Jan 12, 2020 at 15:53
  • \$\begingroup\$ @ThePhoton oh, I see why my method was wrong... but I still don't quite get how to approach the problem then. It seems that now I need to find what inputs were there given the outputs, but what information from the test cases should I use then? \$\endgroup\$ Jan 12, 2020 at 15:59
  • \$\begingroup\$ from Case 1, I can get R1 = 2/4m and from Case 2, I get R2 = 10/10m. Since the resistances won't be changing (they don't explicitly say it's a varying resistor), given I1 and V2, I could solve for V1 and I2, getting V1 = 4 volts and I2 = 4 mA. Is this way of thinking right? \$\endgroup\$ Jan 12, 2020 at 16:06
  • \$\begingroup\$ (I'm afraid that's a bit too simple, actually) \$\endgroup\$ Jan 12, 2020 at 16:06
  • \$\begingroup\$ if the resistor is linear, that is explicitly telling you the resistance value doesn't depend on the applied voltage or current. But what are R1 and R2? There are no such elements mentioned in the problem. \$\endgroup\$
    – The Photon
    Jan 12, 2020 at 16:17

1 Answer 1

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from cases 1 and 2, I should have I1 = 4 + 5 = 9 mA, but it's 8mA.

4 mA + 5 mA is what you would get if you applied 2 V on the left and -10 mA on the right.

But that isn't what you were asked. You were asked what happens if you apply 4 V on the left and 8 mA on the right.

So, to use superposition, first use test result 1 to find out what happens if you apply 4 V on the left (with the right port open). Then use test result 2 to find out what happens if you apply 8 mA on the left. Then sum those results to get your final answer.

Hence, I thought that the black box consumes 1mA;

The black box contains only linear elements. Therefore it can't "consume" any current.

Edit

OK, let me spell things out.

When they say the system is linear, it means you can write these equations for the unknown variables

$$I_1 = G_{11} V_1 + A_{12} I_2$$ $$V_2 = A_{21} V_1 + R_{22} I_2$$

where \$G_{11}\$, \$A_{12}\$, \$A_{21}\$ and \$R_{22}\$ are characteristic parameters of the black box that you need to discover. The names of the parameters are arbitrary, but you should think about why I might have named them what I did.

Your two test measurements give you enough information to determine the unknown parameters.

Then you can plug in any \$V_1\$ and \$I_2\$ you like (for example, the ones you were asked to solve for), and find the resulting \$I_1\$ and \$V_2\$ (for example, the result you were asked to find).

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  • \$\begingroup\$ then is my approach as explained in my comments above closer to what you're telling me? \$\endgroup\$ Jan 12, 2020 at 16:15
  • \$\begingroup\$ "So, to use superposition, first use test result 1 to find out what happens if you apply 4 V on the left (with the right port open)." that does give me 8mA (as was the final test case asked). \$\endgroup\$ Jan 12, 2020 at 16:19
  • \$\begingroup\$ applying 8 mA on the left in Case 2 gives me 8V, which is double the given V2, so I2 = 4 mA as well? \$\endgroup\$ Jan 12, 2020 at 16:20
  • \$\begingroup\$ @singularity, from test 1, 2 V applied on the left gave 8 V out on the right. So applying 4 V will not give 8 V out. It will give a different value that you can figure out because you know the system is linear. \$\endgroup\$
    – The Photon
    Jan 12, 2020 at 16:21
  • \$\begingroup\$ And case 2 applied -10 mA on the right, not +8 mA on the left. \$\endgroup\$
    – The Photon
    Jan 12, 2020 at 16:21

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