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I'm new here and I would like to know if anyone here can help with this question below.

Which resistor must be connected to the circuit of the following image to transfer to it the maximum power?

image of question

My attempt:

my attempt

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  • \$\begingroup\$ Hint: Find the Thevenin equivalent resistance. \$\endgroup\$
    – Barry
    Oct 27, 2020 at 21:17
  • \$\begingroup\$ Another hint - Maximum power transfer occurs when the load impedance equals the source impedance. You could determine that analytically, or maybe by trying a bunch of numbers (in excel or such), or you can just remember that general rule (which your professor apparently hasn't yet shared with you....) \$\endgroup\$
    – Kyle B
    Oct 27, 2020 at 21:37

2 Answers 2

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This is super simple; It's 12kOhm.

Why?

The 18kOhm resistor is in parallel with a voltage source, ie. it has NO influence on anything, throw it away..

The 6kOhm resistor is in series with a current source ie. it has NO influence on anything, throw it away..

What are you left with?

And so what is the output impedance of your circuit???

To get maximum power transfer you just need the load to have the same value as the output impedance of the circuit.

This should be easy enough for you to solve now. (He did solve it)

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  • \$\begingroup\$ so 12kohms. and then the V of the circuit is 8V and not 30V as I thought... \$\endgroup\$
    – Ognum
    Oct 27, 2020 at 21:44
  • \$\begingroup\$ I didn't see anything about a voltage in your question so I didn't consider it, but you are correct that the load needs to be 12kOhm. I have updated my answer. \$\endgroup\$
    – user173292
    Oct 27, 2020 at 21:47
  • \$\begingroup\$ @Ognum and BTW no, the voltage "V" is not 8V. \$\endgroup\$
    – user173292
    Oct 27, 2020 at 21:49
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    \$\begingroup\$ @Ognum if you keep trying I'm sure you'll get through the entire list of options ;) ;) ;). No it is not 13/3mA either.. Look we were all new at this at one point and when I was just two weeks into it I had difficulties with this too, I'm just saying that you need to pay attention to this and study it hard because it is basic and essential to everything in electronics. In other words you should not be so concerned with getting the correct multiple-choice answer you should be concerned with getting the understanding down, screw what results you get at a test, it's understanding that matters. \$\endgroup\$
    – user173292
    Oct 27, 2020 at 22:55
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    \$\begingroup\$ If I could interject, I would suggest that on occasions like this it can be useful to use a circuit simulator to generate the values for you so you can check AFTER you've done the calculations. This would save you from running around in circles and asking Vinzent for every value, though I'm sure they don't mind. LTSpice or Multisim would be a good option if you're at University. \$\endgroup\$
    – ChrisD91
    Oct 27, 2020 at 23:25
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First, I will present a method that uses Mathematica to solve this problem because you already have a good answer by @Vinzent. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_\text{i}=\text{I}_1+\text{I}_2\\ \\ \text{I}_4=\text{I}_\text{k}+\text{I}_2\\ \\ \text{I}_\text{i}=\text{I}_1+\text{I}_3\\ \\ \text{I}_4=\text{I}_\text{k}+\text{I}_3 \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_2}\\ \\ \text{I}_\text{k}=\frac{\text{V}_2-\text{V}_1}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_1}{\text{R}_4} \end{cases}\tag2 $$

Now, we can set-up a Mathematica-code to solve for all the voltages and currents:

In[1]:=FullSimplify[
 Solve[{Ii == I1 + I2, I4 == Ik + I2, Ii == I1 + I3, I4 == Ik + I3, 
   I1 == Vi/R1, I2 == (Vi - V1)/R2, Ik == (V2 - V1)/R3, 
   I4 == V1/R4}, {Ii, I1, I2, I3, I4, V1, V2}]]

Out[1]={{Ii -> Vi/R1 + (-Ik R4 + Vi)/(R2 + R4), I1 -> Vi/R1, 
  I2 -> (-Ik R4 + Vi)/(R2 + R4), I3 -> (-Ik R4 + Vi)/(R2 + R4), 
  I4 -> (Ik R2 + Vi)/(R2 + R4), V1 -> (R4 (Ik R2 + Vi))/(R2 + R4), 
  V2 -> (Ik R2 R3 + Ik (R2 + R3) R4 + R4 Vi)/(R2 + R4)}}

Now, we can find:

$$\text{P}_{\text{R}_4}=\text{V}_{\text{R}_4}\cdot\text{I}_{\text{R}_4}=\text{V}_1\cdot\text{I}_4=\text{R}_4\cdot\left(\frac{\text{V}_\text{i}-\text{I}_\text{k}\text{R}_2}{\text{R}_2+\text{R}_4}\right)^2\tag3$$

So, we want to find \$\frac{\partial\text{P}_{\text{R}_4}}{\partial\text{R}_4}=0\space\Longleftrightarrow\space\text{R}_4=\dots\$. In order to solve it, I used the following Mathematica-code:

In[3]:=FullSimplify[
 Solve[{D[(R4 (Ik R2 + Vi))/(R2 + R4)*(Ik R2 + Vi)/(R2 + R4), R4] == 
    0, R4 > 0 && Ik > 0 && Vi < 0 && R2 > 0}, R4]]

Out[3]={{R4 -> ConditionalExpression[R2, Vi < 0 && R2 > 0 && Ik > 0]}}

Using your values we get:

$$\text{R}_4=\text{R}_2=12\space\text{k}\Omega\tag6$$

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