Electrical Network Problem: Power Transfer

[cnh1995]

Homework Statement

Homework Equations

P=V²/R
For maximum power transfer,
R_load=R_Thevenin

The Attempt at a Solution

I know the efficiency is 50% during maximum power transfer. So when all the sources are acting simultaneously and ( at the same time) R=R_th, maximum power will be dissipated in R.
I am not sure how to use the given values of power since both R and the source voltages are unknown. Plus, the voltage sources are connected across three different ports. This problem is from Circuit Theorems chapter, so I guess I need to use Maximum power transfer theorem along with Superposition theorem.
Any guidance is appreciated.

 

[haruspex]

For maximum power transfer,
Rload=RThevenin

I don't think this is relevant here. That equation says how to adjust the load to get maximum power transfer into it.
Consider that the direction of current might change between the three given power results.

 

[cnh1995]

Consider that the direction of current might change between the three given power results.

Yes, but I don't know how to relate the three power results. I can't directly add/subtract the power results since I don't know the network inside the box. I think I need to set up an equation using some sort of proportionality (P∝V² for instance).
The answer given is 450W (max) and 2W (min).
Do I need to use any specific network theorem here?

 

[cnh1995]

Yes, but I don't know how to relate the three power results. I can't directly add/subtract the power results since I don't know the network inside the box. I think I need to set up an equation using some sort of proportionality (P∝V² for instance).
The answer given is 450W (max) and 2W (min).
Do I need to use any specific network theorem here?

Plus, when they say 'one source is acting at a time', do I open-circuit the other two sources or short them?
I now doubt if this problem is solvable with the given data.

 

[gneill]

Plus, when they say 'one source is acting at a time', do I open-circuit the other two sources or short them?

Presumably they mean to suppress the sources as is done in superposition analysis. So, short them since they are voltage sources.

Suppose that each individual source produces some potential difference across R. By superposition you know that these potential differences will sum linearly, and the resulting power will go as the square of that sum. But you are given individual power dissipations instead. How might you sum them?

 

[cnh1995]

Presumably they mean to suppress the sources as is done in superposition analysis. So, short them since they are voltage sources.

Suppose that each individual source produces some potential difference across R. By superposition you know that these potential differences will sum linearly, and the resulting power will go as the square of that sum. But you are given individual power dissipations instead. How might you sum them?

If E₁, E₂ and E₃ produce voltages V₁, V₂ and V₃ respectively across R individually, then in combined case,
P=V²/R,
where V=aV₁+bV₂+cV₃.
Is that right?

 

[gneill]

If E₁, E₂ and E₃ produce voltages V₁, V₂ and V₃ respectively across R individually, then in combined case,
P=V²/R,
where V=aV₁+bV₂+cV₃.
Is that right?

Yes. You can also say that the individual potential differences expressed across R will be proportional to the square root of the individual powers.

 

[haruspex]

If E₁, E₂ and E₃ produce voltages V₁, V₂ and V₃ respectively across R individually, then in combined case,
P=V²/R,
where V=aV₁+bV₂+cV₃.
Is that right?

That looks reasonable, but there are limits on a, b and c.

 

[Charles Link]

As per post 7, given the termination has some $R_o$, $P_a=\frac{V_a^2}{R_o}$, $P_b=\frac{V_b^2}{R_o}$, and $P_c=\frac{V_c^2}{R_o}$. A little algebra gives $\sqrt{P_a}+\sqrt{P_b}+\sqrt{P_c}=\frac{V_a+V_b+V_c}{\sqrt{R_o}}=\frac{V_{total}}{\sqrt{R_o}}$. It is a simple matter to compute $P_{total}$ by squaring both sides. $\\$ Edit: One additional item though: What if the sign of $V_b$ or $V_c$ is different from that of $V_a$? The above equality doesn't always hold. That's apparently why there is a maximum and a minimum. $\\$ Additional edit: And yes, I get the answers of post 3 by selecting the appropriate relative signs for the 3 terms.

 

[cnh1995]

Sorry for replying so late!

Thanks a lot @gneill, @Charles Link and @haruspex for your clever hints. Turns out what I really need here is only the Superposition theorem. I got the correct answer now.
Thanks again!