Question from exam, couldn’t do this question and dont understand how to…

The question was something like:
Determine the value of resistor R so that the maximum power can be delivered to the resistor. Compute the maximum power delivered.
Thanks in advance, will have trouble sleeping until I understand this.

This question requires that you use a combination of node and mesh analysis as described in this document
https://ocw.mit.edu/courses/electrical-engine...pring-2006/lecture-notes/nodal_mesh_methd.pdf
The tricky thing about the circuit in your question is that you have a dependent voltage source on one side and a current source on the other. It's not obvious what the best approach is for this circuit.I'd just take a brute force approach and write down all the node and loop equations then manipulate them to get an equation for the power in R.Hmm, then you'd need to differentiate to find the maximum... This is way harder than it first appears....

Elizabeth is right, this is a stinker of a question.  But my attitude to these questions is, they bear no relation to anything you will find in the real world.  So don't lose sleep over this.  There are more important things to worry about.  So unless your future depends on it, don't worry too much about it....

Could try another approach, maximum power occurs in the load when it is equal to the source resistance driving it, cheers,Richard

By dimensional analysis, the "2io" source on the left side must be a current source. For it to be a voltage source, the label would have to be "2io*some_resistance". Of course, the simulator may use naked unitless numeric values and ignore dimensional consistency, in which case the left source could be a voltage source. A voltage source makes more sense, as the 1-ohm resistor in series with a current source does nothing. I have never used graphical-input circuit simulators (I have only used SPICE text-input circuit simulators), so I am unfamiliar with the details of symbols unique to graphical input.If the left source is a current source, then the currents flowing are independent of the value of R, and the power dissipation in R is proportional to R.
The left source being a voltage source (which should be labeled "2io*1ohm") makes for a more-interesting problem, but the practice of excluding units from quantities is dangerous. NASA lost a Mars probe because of using unitless distances and speeds: Some engineers did not realize correctly whether the units were U.S. or Metric.David Ashton is correct, this is an academic-stinker problem you are unlikely to encounter in the real world.

Thanks for the help! To address some things:I think we can look at the 2*io as voltage and assume that the 2 has the unit, Ohm. I've learned that the maximum power is Pmax = V2Th / 4RTh , so if we can find the Thevenin eq. over R then maximum power is delivered when R = RThBut I have trouble to understand how to get the Thevenin eq.
Thanks again, and I could sleep, but this is still bugging me!

If the source on the right is a 1-ampere current source, then the 1-ohm resistor in series does nothing. Are you assuming that the source on the right is a 1-ampere current source?If the voltage at the junction of the three resistors drops below minus one volt, the 1-ampere current source on the right will be sinking power rather than sourcing power.After you calculate your solution, show it to us. I plan to calculate a solution, and we can compare. Since I never had undergraduate EE classes, I can learn from comparing solutions.

Peter,
The left source is a current dependent voltage source. The diamond shape is typically used to denote dependent sources. For more information, see this tutorial (scroll down to the bottom where they discuss dependent voltage sources.
https://www.electronics-tutorials.ws/dccircuits/voltage-source.htmlThe next page in the series is on current sources.By the way, the current dependent voltage source is often used in small signal analysis of bipolar transistors.

I believe I have a solution. Here is the approach I used.First, Peter is correct that the 1-ohm resistor in series with the 1A current source does nothing.  It's just there to provide confusion.If you choose your reference to be the node at the bottom, it's easy to write equations for the voltage on the left and the voltage in the center in terms of io and R. Then you can write an equation for the current through the resistor on the left.
You now have equations for all the currents so you can write the node equation and simplify to get an equation for R in terms of io.Since we're looking for the power through R, I wrote down the formula for power through a resistor.From there you do substitutions and differentiate to find the max.Notice that I didn't have to find the Thevenin equivalent circuit. Which is a good thing because I'm not sure how to do that with a dependent source (or if it's even possible with this circuit).OK now I can sleep at night at least. LOL

I used Elizabeth's approach, and had  a solution an hour  and a half later. How much time did the exam give to solve this problem?The current-dependent voltage source makes this circuit nonlinear, so Thevenin equivalent is difficult.

After we've solved the DC problem, we could take our analysis a step further and do a stability analysis. The current-controlled voltage source is a transresistance amplifier. The resistors form a feedback loop. Is the feedback positive or negative? For what values of R is the circuit stable, and for what values of R is the circuit unstable?

To double-check that I read the schematic correctly: io and 1A show current arrows pointing up. I interpret the arrows as being conventional positive-charge current flow. Am I correct?If we have an upward conventional current flowing through the voltage source on the left, I think the source is delivering power to the circuit. Am I correct?

Peter,I left the paper with my analysis on my desk at work so I don't have it in front of me but I believe I got a negative result for io (assuming the arrow indicates the positive direction). Just another aspect of this problem designed to confuse the student...

Elizabeth, I got io=-0.5A and R=1ohm, so I am understanding the sign conventions the same way you are.

My further calculations show that for R<1ohm, io is infinite. This implies that, if we put the circuit into a simulator program with R<1ohm, the iterative numerical DC analysis would crash, which would really confuse the student. The crash occurs because the circuit is unstable for R<1ohm, and io would grow with each iteration of the DC solver.

The feedback loop around the current-controlled voltage source, the left-hand 1-ohm resistor, and R, is a positive-feedback loop. For R>1ohm, the loop gain is less than one and the loop is stable. For R<1ohm, the loop gain is more than one and the loop is unstable.

Hmm, I got R = 3ohm so somehow our calculations aren't matching up...It's all too easy on these problems to get a sign swapped when simplifying the equations.

i hope i never see anything like that on the exam.

@josef: "...i hope i never see anything like that on the exam..."One of my friends told me that he was presented with a mind-boggling logic problem during an interview -- the idea is that you have a black box with three inputs A, B, C and three outputs !A, !B, !C where the outputs are the logical inversions of the inputs. The trick is that you can use as many AND and OR gates as you wish, but only TWO NOT gates (and no XOR or XNOR gates). You can read the full problem in my Logic Gate Brain Boggler column, after which you can check out my Brain Boggler Solutions column.

Elizabeth, thanks for finding the sign error in my solution. My solution now agrees with your solution. Also, the loop gain is negative for all positive values of R, so the circuit is stable.