Maximum power transfer problem. Find the resistance

Question

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?

My attempt:

Answer 1

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)

Answer 2

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:

^{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:

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