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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, 2018 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, 2018 at 10:04
  • \$\begingroup\$ Nope, this is the circuit exactly. \$\endgroup\$
    – JustHeavy
    Oct 5, 2018 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, 2018 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, 2018 at 20:52

2 Answers 2

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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, 2018 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, 2018 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, 2018 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, 2018 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, 2018 at 18:01
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Please see update #1 below. My original answer (immediately following this paragraph) assumes that the voltage controlled voltage source in your schematic has a certain polarity. If you read the article Dependent Sources and Thevenin's Theorem, their convention for dependent current source polarity is somewhat counter-intuitive (for me, at least), but applying that convention to your circuit does yield a workable solution, whereas my initial assumption does not, as I describe now.

I spent a couple of hours trying to find the Thevenin equivalent of this circuit, before it dawned on me that this is not necessarily possible. So I attacked the whole circuit with nodal analysis.

schematic

simulate this circuit – Schematic created using CircuitLab

These are the equations I came up with (polarities are critical, so pay attention to the signs):

KVL applied to the 4 loops:

$$ 4 \times I_O + V_X + I_4R = 0V $$

$$ 5V + 4 \times I_O - 1\Omega \times I_1 = 0V $$

$$ 1\Omega \times I_1 + V_X - 4\Omega \times I_O = 0V $$

$$ 4\Omega \times I_O + 2\Omega \times I_3 - 10V = 0V $$

KCL applied to node B (remembering that \$V_O = I_4R\$):

$$ I_3 + I_4 - I_O - 2I_4R = 0A $$

There are five unknowns, \$V_X\$, \$I_O\$, \$I_1\$, \$I_3\$ and \$I_4\$, and five simultaneous equations. Solving for \$I_4\$, these whittle down to the following relationship between resistance R and the current through it, \$I_4\$:

$$ I_4 = -\frac{5}{4-5R} $$

Without any differentiation, or any use of the Maximum Power Transfer Theorem, it's quite clear that the denominator approaches zero as R approaches 800mΩ. That's where current is at an impossible maximum of ∞A, and where power will be at a maximum.

Are you sure you copied the schematic correctly? Maybe I didn't.

Here's a working CircuitLab model, where you can verify this discontinuity:

schematic

simulate this circuit

Update #1

Until now I assumed that the current source produced current to the left when \$V_O\$ is positive, which may not be the case. Here I perform a complete nodal analysis for the case where we reverse this current direction, but this is only one way to find the maximum power in R. A better way may be to find the Thevenin equivalent circuit between A and B, which I'll do after. The equations become:

KVL around the four loops:

$$ 4 \times I_O + V_X + I_4R = 0V $$

$$ 5V + 4 \times I_O - 1\Omega \times I_1 = 0V $$

$$ 1\Omega \times I_1 + V_X - 4\Omega \times I_O = 0V $$

$$ 4\Omega \times I_O + 2\Omega \times I_3 - 10V = 0V $$

KCL (again keeping in mind that \$V_O = I_4R\$):

$$ I_3 + I_4 - I_O \overbrace{+ 2I_4R}^{\text{sign change}} = 0A $$

This time, solving for \$I_4\$ reveals this relationship, which does not suffer the same discontinuity:

$$ I_4 = -\frac{5}{4+11R} $$

That means, as far as I can tell, that there does exist a Thevenin equivalent, which I'll derive in moment. For now, I want to find algebraically the value of R which will dissipate the most power, which requires the power law:

$$ \begin{aligned} P &= I_4^2 R \\ \\ &= (\frac{-5}{4+11R})^2 R \\ \\ &= \frac{25R}{121R^2 + 88R + 16} \end{aligned} $$

I'll cheat with Wolfram Alpha to find the derivative, and set it equal to zero (to find maxima and minima):

$$ \frac{dP}{dR} = 0 = -\frac{25(11R-4)}{(11R+4)^3} $$

Wolfram Alpha also provides the solution, which is:

$$ R = \frac{4}{11}\Omega = 363.6m\Omega $$

To find the Thevenin equivalent circuit, we need to know the open circuit voltage, which looks like this, and where we need to find \$V_O\ = V_A-V_B\$:

schematic

simulate this circuit

The KVL and KCL equations for this are the same, with a couple of changes:

  1. I'll replace \$I_4R\$ with \$V_O\$, the voltage between A and B which we wish to find, and
  2. \$I_4 = 0\$, since there's no resistor to draw current from \$V_O\$.

$$ 4 \times I_O + V_X + V_O = 0V $$

$$ 5V + 4 \times I_O - 1\Omega \times I_1 = 0V $$

$$ 1\Omega \times I_1 + V_X - 4\Omega \times I_O = 0V $$

$$ 4\Omega \times I_O + 2\Omega \times I_3 - 10V = 0V $$

$$ I_3 - I_O + 2V_O = 0A $$

To facilitate the solution, using matrix manipulations, or inverse, or whatever technique you want to use, here are those same equations with all the unknowns in columns:

$$ \begin{array}{lllllll} &+V_O &+V_X &+4I_O & & &= &0 \\ & & &+4I_O &-I_1 & &= &-5 \\ & &+V_X &-4I_O &+I_1 & &= &0 \\ & & &+4I_O & &+2I_3 &= &+10 \\ &+2V_O & &-I_O & &+I_3 &= &0 \\ \end{array} $$

I'll cheat again, and use an online solver, which gave me this result:

$$ \begin{aligned} V_O &= -\frac{5}{11} &= -454.5mV \\ \\ V_X &= -5V \\ \\ I_O &= +\frac{15}{11} &= +1.364A \\ \\ I_1 &= +\frac{115}{11} &= +10.46A \\ \\ I_3 &= +\frac{25}{11} &= +2.273A \end{aligned} $$

Our Thevenin voltage is \$V_{TH} = V_O = -454mV\$.

We can find Thevenin resistance \$R_{TH}\$ by finding short circuit current. Since we already have the relationship between \$I_4\$ and R, this is trivial. Simply set R to zero:

$$ I_4 = -\frac{5}{4+11R} = -\frac{5}{4} = -1.250A $$

From there, finding \$R_{TH}\$ is a case of finding what resistance across \$V_{TH}\$ would produce -1.25A:

$$ \begin{aligned} R_{TH} &= \frac{V_{TH}}{I_4} \\ \\ &= \frac{-454.5mV}{-1.250A} \\ \\ &= 363.6m\Omega \end{aligned} $$

However, that approach assumes you've derived the formula relating \$I_4\$ and R, and frankly that was a pain to do. So perhaps it would be easier to modify those simultaneous equations one more time, to find short circuit current via a matrix solution instead.

Firstly, we set \$V_O = I_4R = 0V\$ in all the original equations:

$$ 4 \times I_O + V_X + 0V = 0V $$ $$ 5V + 4 \times I_O - 1\Omega \times I_1 = 0V $$ $$ 1\Omega \times I_1 + V_X - 4\Omega \times I_O = 0V $$ $$ 4\Omega \times I_O + 2\Omega \times I_3 - 10V = 0V $$ $$ I_3 + I_4 - I_O + 0A = 0A $$

As a pseudo-matrix:

$$ \begin{array}{lllllll} &+V_X &+4I_O & & & &= &0 \\ & &+4I_O &-I_1 & & &= &-5 \\ &+V_X &-4I_O &+I_1 & & &= &0 \\ & &+4I_O & &+2I_3 & &= &+10 \\ & &-I_O & &+I_3 &+I_4 &= &0 \\ \end{array} $$

The solutions are:

$$ \begin{aligned} V_X &= -5V \\ \\ I_O &= +\frac{5}{4} &= +1.250A \\ \\ I_1 &= +10A \\ \\ I_3 &= +\frac{5}{2} &= +2.500A \\ \\ I_4 &= -\frac{5}{4} &= -1.250A \end{aligned} $$

That's the same result for \$I_4\$ short-circuit current as we predicted before, and will of course yield the same \$R_{TH}\$. By the Maximum Power Transfer Theorem, clearly the load resistor R should be \$ R = R_{TH} = 363.6m\Omega \$ for it to dissipate maximum power, which agrees with the algebraic result we found earlier.

Here's a working CircuitLab model to see for yourself that it's all good. Notice that I've reversed the + and - connections on the current source, to correct the current direction, and resolve the discontinuity problem I encountered when I originally answered this question:

schematic

simulate this circuit

Here's a graph of power in R vs. R, to see the peak power at 363.6mΩ:

enter image description here

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