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

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#1 21681192 28 Nov 2018 18:54

Olof Holmsen Olof Holmsen

Anonymous
Post #1
21681192 28 Nov 2018 18:54

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.

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#2 21681193 28 Nov 2018 23:19

Elizabeth Simon Elizabeth Simon

Anonymous

Post #2
21681193 28 Nov 2018 23:19

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....
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#3 21681194 29 Nov 2018 01:53

David Ashton David Ashton

Anonymous

Post #3
21681194 29 Nov 2018 01:53

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....
#4 21681195 29 Nov 2018 04:34

Richard Gabric Richard Gabric

Anonymous

Post #4
21681195 29 Nov 2018 04:34

Could try another approach, maximum power occurs in the load when it is equal to the source resistance driving it, cheers,Richard
#5 21681196 29 Nov 2018 07:22

PeterTraneus Anderson PeterTraneus Anderson

Anonymous

Post #5
21681196 29 Nov 2018 07:22

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.
#6 21681197 29 Nov 2018 10:58

Olof Holmsen Olof Holmsen

Anonymous

Post #6
21681197 29 Nov 2018 10:58

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!
#7 21681198 30 Nov 2018 00:34

PeterTraneus Anderson PeterTraneus Anderson

Anonymous

Post #7
21681198 30 Nov 2018 00:34

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.
#8 21681199 30 Nov 2018 06:28

Elizabeth Simon Elizabeth Simon

Anonymous

Post #8
21681199 30 Nov 2018 06:28

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.
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#9 21681200 30 Nov 2018 07:07

Elizabeth Simon Elizabeth Simon

Anonymous

Post #9
21681200 30 Nov 2018 07:07

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
#10 21681201 30 Nov 2018 08:38

PeterTraneus Anderson PeterTraneus Anderson

Anonymous

Post #10
21681201 30 Nov 2018 08:38

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.
#11 21681202 01 Dec 2018 03:56

PeterTraneus Anderson PeterTraneus Anderson

Anonymous

Post #11
21681202 01 Dec 2018 03:56

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?
#12 21681203 02 Dec 2018 06:50

PeterTraneus Anderson PeterTraneus Anderson

Anonymous

Post #12
21681203 02 Dec 2018 06:50

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?
#13 21681204 02 Dec 2018 09:59

Elizabeth Simon Elizabeth Simon

Anonymous

Post #13
21681204 02 Dec 2018 09:59

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...
#14 21681205 02 Dec 2018 11:12

PeterTraneus Anderson PeterTraneus Anderson

Anonymous

Post #14
21681205 02 Dec 2018 11:12

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.
#15 21681206 04 Dec 2018 00:26

Elizabeth Simon Elizabeth Simon

Anonymous

Post #15
21681206 04 Dec 2018 00:26

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.
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#16 21681207 04 Dec 2018 21:28

Josef Darling Josef Darling

Anonymous

Post #16
21681207 04 Dec 2018 21:28

i hope i never see anything like that on the exam.
#17 21681208 05 Dec 2018 03:39

Max Maxfield Max Maxfield

Anonymous

Post #17
21681208 05 Dec 2018 03:39

@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.
#18 21681209 21 Dec 2018 08:43

PeterTraneus Anderson PeterTraneus Anderson

Anonymous

Post #18
21681209 21 Dec 2018 08:43

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.
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Topic summary

✨ The discussion centers on determining the resistor value R that maximizes power delivery in a circuit containing a current-dependent voltage source and a current source. The problem is complex due to the presence of dependent sources and feedback loops, making Thevenin equivalent analysis challenging. Approaches suggested include using node and mesh analysis to write and solve circuit equations, differentiating power expressions to find maxima, and applying the maximum power transfer theorem where maximum power occurs when the load resistance equals the source (Thevenin) resistance. The circuit features a current-controlled voltage source (transresistance amplifier) and a 1-ampere current source in series with a 1-ohm resistor, which is effectively negligible. Stability analysis reveals a positive feedback loop involving the dependent source and resistors, with circuit stability dependent on R values; specifically, the circuit is stable for R > 1 Ω and unstable for R < 1 Ω. Sign conventions and correct interpretation of dependent sources are critical, as errors can lead to contradictory or unstable solutions. The problem is noted as difficult and somewhat artificial compared to practical real-world circuits.

FAQ

TL;DR: For this CCVS exam circuit, the max-power condition evaluates to R = 1 Ω; “io=-0.5A and R=1ohm.” [Elektroda, Anonymous, post #21681205]

Why it matters: If you’re revising how to pick R for maximum power with dependent sources, this FAQ shows the steps and pitfalls clearly—ideal for students and self-learners.

Quick Facts

Final thread result: R = 1 Ω and io = −0.5 A at the operating point. [Elektroda, Anonymous, post #21681205]
Diamond symbol denotes a dependent source; here it’s a current-controlled voltage source (CCVS). [Elektroda, Anonymous, post #21681199]
The 1 Ω in series with a 1 A current source doesn’t affect DC; it’s distraction. [Elektroda, Anonymous, post #21681200]
Dependent source makes the circuit nonlinear, complicating Thevenin use. [Elektroda, Anonymous, post #21681201]
Edge case: if the center node drops below −1 V, the right current source sinks power. [Elektroda, Anonymous, post #21681198]

What exactly was the exam-style question here?

Determine R so the resistor receives maximum power, then compute that maximum power. The circuit mixes a current-controlled voltage source on the left and a 1 A source on the right, which makes analysis nontrivial. [Elektroda, Anonymous, post #21681193]

How do I pick R for maximum power in this circuit?

Use nodal or mesh analysis to express power in R as a function of R, then differentiate and set dP/dR=0. A participant concluded R = 1 Ω with io = −0.5 A after solving. “Way harder than it first appears,” but tractable. [Elektroda, Anonymous, post #21681205]

Can I just use the Thevenin maximum-power rule R = RTh?

Yes in principle, but extracting RTh with a CCVS is awkward. One poster instead wrote node equations, found P(R), and optimized directly. That avoided building a Thevenin model around a nonlinear dependent source. [Elektroda, Anonymous, post #21681200]

What final values did the solvers in the thread agree on?

They reconciled to io = −0.5 A and R = 1 Ω. This aligns with the computed operating point once sign conventions were corrected. [Elektroda, Anonymous, post #21681205]

Is the loop stable for positive R?

A later correction notes the loop gain is negative for all positive R, so the circuit is stable in that region. That addresses earlier concerns about instability claims. [Elektroda, Anonymous, post #21681209]

Is the left diamond source a voltage or current source?

It is a current-controlled voltage source (CCVS). The diamond marks dependency; the source’s voltage is proportional to a measured current, matching small-signal transistor modeling practice. [Elektroda, Anonymous, post #21681199]

Does the 1 Ω series resistor with the 1 A source matter?

No for DC analysis here. A 1 Ω in series with an ideal current source does not change the enforced 1 A; it’s present to distract. [Elektroda, Anonymous, post #21681200]

Why do folks say this question is difficult?

You must mix nodal/mesh methods and handle a dependent source. As one expert put it, it’s “way harder than it first appears,” especially under exam timing. [Elektroda, Anonymous, post #21681193]

How should I solve it step by step without Thevenin?

Choose the bottom node as reference and write node voltages in terms of io and R.
Write KCL at the center node; express all currents via node voltages.
Substitute into P=V^2/R or I^2R for the load, differentiate P(R), set derivative to zero. [Elektroda, Anonymous, post #21681200]

What happens if the center node voltage drops below −1 V?

The right-hand 1 A source would absorb power instead of supplying it. That sign flip is an important edge case when checking power directions. [Elektroda, Anonymous, post #21681198]

Do current direction arrows define actual current direction?

They define reference direction. If your solved current is negative, actual current flows opposite the arrow. One solver found io negative, consistent with the diagram. [Elektroda, Anonymous, post #21681204]

Is there a quick heuristic for max power if I did have a Thevenin model?

Yes. For a linear Thevenin source, max power in the load occurs when Rload equals RTh. This rule of thumb guided one poster’s thinking before full analysis. [Elektroda, Anonymous, post #21681195]

Where can I review nodal and mesh methods fast?

See the referenced MIT OCW notes covering nodal and mesh analysis with worked steps. They’re concise and ideal for exam prep. “Nodal and Mesh Methods (MIT OCW)”

Where can I read a refresher on dependent sources?

The linked tutorial explains dependent voltage and current sources and their symbols, with CCVS examples you can mirror in hand analysis. [Electronics-Tutorials.ws]

What was the consensus about using simulators on this circuit?

Participants emphasized understanding first. After reconciling signs, the hand solution was consistent, and no special simulator tricks were required. [Elektroda, Anonymous, post #21681209]