Finding the missing current and voltage in a circuit with superposition

[JessicaHelena]

Homework Statement

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.

Relevant Equations

superposition?
V/I = R? (R doesn't really concern us here, I think)

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.

 

[TSny]

Suppose the variable $z$ is a linear superposition of the variables $x$ and $y$. That doesn't necessarily mean that $z = x+y$. It means that there exist constants $a$ and $b$ such that $z = ax + by$.

You might want to make a table for the three cases that you are working with:

$$\,\,\, \underline{ V_1 \,\,\, I_1 \,\,\, V_2 \,\,\, I_2} \\ \,\,\,\, 2 \,\,\,\,\,\, 4 \,\,\,\,\, 8 \,\,\,\,\, 0 \\
\,\,\,\, \cdot \,\,\,\,\,\,\cdot \,\,\,\,\, \cdot \,\,\,\,\,\cdot \\
\,\,\,\, \cdot \,\,\,\,\,\,\cdot \,\,\,\,\, \cdot \,\,\,\,\,\cdot $$

I already filled in the first row for the first situation. Fill in the other two rows as much as you can for the second and third situations. Then, try to find values of $a$ and $b$ that makes the third row a linear superposition of the first two rows.

 

[JessicaHelena]

@TSny
Oh, so letting x be the first row, y be the second row, and z be the third row, I have that 4a + 5b = 8 and 8a - 10b = 4. Solving this system, a = 5/4 and b = 3/5.

Since V1 in Z is 2a + 0b, it's 5/2 and I2 in z = 0a - 10b = -6 mA. Is this right?

 

[TSny]

I believe that's right.

 

[JessicaHelena]

Thank you so much!

 

[Babadag]

In two-port network with only passive elements and only linear elements one may use these equations:

V1=Z11*I1+Z12*I2

V2=Z21*I1+Z22*I2

V1,V2,I1,I2 change but Z11,Z12,Z21,and Z22 do not.

Then in first case in which I2=0 we apply V1 and get I1,I2 and V2.From the above equation we can get:

V1=Z11*I1 and Z11=V1/I1

V2=Z21*I1 then Z21=V2/I1

In second case [V1=0] we supply I2 and already knowing Z11 and Z21 we may calculate Z12 and Z22 from:

0=Z11*I1+Z12*I2

V2=Z21*I1+Z22*I2

Finally supplying V1 and I2 [unknown yet] we get V2 and I1 then:

V2=Z21*I1+Z22*I2

I2=(V2-Z21*I1)/Z22 and V1=Z11*I1+Z12*I2