Tellegen's Theorem Question

Question

In Irwin's textbook on Basic Circuit Analysis, there is the following exercise:

Up to this point in the book, the concepts of dependent and independent sources have been discussed, and also the issues related to power supply and power consumption (\$p = vi\$).

Kirchhoff's Laws have not been discussed.

Here two questions come to mind.

How to know the current in component 2, since the closest information is about the current \$I_x\$, which is given by an independent current source.

Is the question missing some data?

The answer key mentions that the result is \$I_x = 2\$ A. My first thought to solve this with Tellegen's Theorem (\$\sum_j p_j = 0\$), but for that I would need these two pieces of information above, correct?

Answer 1

(I decided not to wait longer. A couple of hours is long enough.)

You can use Tellegen's. And it doesn't matter how you decide to estimate the power in each branch shown. You can decide that a current moving up is positive, or negative, and you can decide that a voltage with the + on top is positive, or negative. Doesn't matter. Always get the same answer.

Let's start by assuming that the power is equal to the top-to-bottom voltage difference times the current, where a current moving from top-to-bottom is also positive. All the branches, including the Box 1 branch, must have \$+25\:\text{V}\$ across them. So in this case, you have:

$$+25\:\text{V}\cdot\left(1\:\text{A}-5\:\text{A}-I_x+2\:\text{A}\right)=0\:\text{W}$$

(I've collected the common voltage to the left, above. But feel free to distribute it across the currents so that you cast it directly in terms of Tellegen's. \$V_i=V\$, as \$V\$ is common to all of the branches. So collecting it to the left is fine in order to simplify \$\sum V_i\cdot I_i = V\sum I_i=0\:\text{W}\$.)

That solves as \$I_x=-2\:\text{A}\$. If you assume that a positive current is one that goes from bottom-to-top then:

$$+25\:\text{V}\cdot\left(-1\:\text{A}+5\:\text{A}+I_x-2\:\text{A}\right)=0\:\text{W}$$

That solves as \$I_x=-2\:\text{A}\$.

And I don't suppose I need to tell you that if you decided that the voltage across each branch was instead negative, that you would still get the same result: \$I_x=-2\:\text{A}\$.

Of course, KCL would also tell you this. If \$ V\sum I_i=0\:\text{W}\$ and \$V\ne 0\:\text{V}\$ then \$\sum I_i=0\:\text{A}\$, which is KCL. But if all you have is Tellegen's and the \$P=V\cdot I\$ formula, then you are still fine.