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This relates to max power transfer. Here's my circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

My goal is to try and find the value of R for the maximum power that can be delivered to it. Then I'll calculate that power as well. I have tried drawing the circuit with an a port on the left of the R resistor and a b port on the right of the R resistor and then trying to calculate Rth by shorting the independent voltage sources, but each time I do that and apply KCL or KVL I get equations that equal zero or not enough equations to solve for all the unknowns.

That's just Rth, I do not even know where to get started for calculating Vth for this circuit. My best guess is that I will ignore the R or load resistor and then apply nodal analysis or some other techniques but I'm not sure which one to use.

I know these equations: $$V_{oc}=V_{th}\;\;\;\;\;\;I_N=I_{sc}\;\;\;\;\;\;R_N=R_{th}={V_{oc}\over I_{sc}}\;\;\;\;\;\;R_L=R_{th}\;\;\;\;\;\;p_{max}={V_{th}^2 \over {4R_{th}}}\;\;\;\;\;\;When\;\;\;R_L \ne R_{th}\;\;\;p=i^2R_L=({V_{th}\over{R_{th}+R_L}})^2R_L$$ I know those equations but just do not know how to get them correctly.

Sorry I can't show work or explain it more, this is as far as I can get right now. Any help would be very appreciated. Thank you!

Right now I'm most familiar with superposition, Thevenin, Norton, Nodal, Mesh, and Ohm's law.

EDIT: in case anyone is still reading this, I'm still stuck and every time I try to solve for Voc or Isc I get a system of equations that can't be solved.

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  • \$\begingroup\$ I'd rather try to get Thevenin equivalent through short circuit current and open circuit voltage at R port.. you'll get two simple to solve circuits. It is always good idea to cut the problems into smaller chunks, brute force approach is for computers instead. \$\endgroup\$ – carloc Oct 5 '18 at 8:53
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    \$\begingroup\$ I don't see how the current dependent voltage source or the \$1\:\Omega\$ resistor means anything to the answer. You could just drop those out of the circuit and ground that now-loose end of the remaining voltage dependent current source and still have the same problem to solve. (I suspect you didn't transcribe the circuit well. Perhaps something isn't right?) \$\endgroup\$ – jonk Oct 5 '18 at 10:04
  • \$\begingroup\$ Nope, this is the circuit exactly. \$\endgroup\$ – JustHeavy Oct 5 '18 at 18:11
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    \$\begingroup\$ @JustHeavy I'm trying to get you to see something here that makes this circuit somewhat unique (in a mathematical sense.) Given the other comments here, I am not sure others who've written yet "see" the issue. They seem to imagine that if you follow a basic process the end of that path with give you something reasonable. Actually, this is an interesting question now that I look more closely. \$\endgroup\$ – jonk Oct 5 '18 at 20:44
  • \$\begingroup\$ Thanks for your input, but from what I have learned so far I can't see how I could just 'lose' the dependent source. I'm starting to think I can't find the value R. I have put too many hours into this problem now. \$\endgroup\$ – JustHeavy Oct 5 '18 at 20:52
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Use Thevenin, with \$\small R\$ as the load. Determine \$\small R_{TH}\$ (turn all sources off); then maximum power transfer will be when \$\small R=R_{TH}\$. No need to determine \$\small V_{TH}\$ or \$\small I_{SC}\$.

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  • \$\begingroup\$ Either Isc or Vth is needed to calculate the power as requested, but their are also the simplest way to get Rth. \$\endgroup\$ – carloc Oct 5 '18 at 9:57
  • \$\begingroup\$ @carloc \$\small R_{TH}\$ is simply 2||4 ohms. The question appears to be to find R, and '... then I'll calculate the power as well...' looks like an afterthought. \$\endgroup\$ – Chu Oct 5 '18 at 10:37
  • \$\begingroup\$ Are you sure the VCCS at node V3 rules out from Rth? For the afterthought you may be right though \$\endgroup\$ – carloc Oct 5 '18 at 10:46
  • \$\begingroup\$ @carloc, I don't know what V3 is; I took it as just a random, like the I0 and the clumsy -> arrow! \$\endgroup\$ – Chu Oct 5 '18 at 10:50
  • \$\begingroup\$ You're correct. The V3 and I0 are just given in the circuit drawing, they are not needed. I do wish to calculate the power but that should be simple when the variables are found. \$\endgroup\$ – JustHeavy Oct 5 '18 at 18:01

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