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Someone sent me this question along with an answer. However I am completely unable to calculate this using either KCL or KVL.

schematic

simulate this circuit – Schematic created using CircuitLab

The value for R is 10 ohm for maximum power transfer. But as per my calculation maximum power transfer will be on a value Rinfinity. (With R increasing V1 increases which in turn will increase current 0.1V1)

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  • \$\begingroup\$ Maybe you can share your calculation so we can see where it went wrong? With two separate sources you should use superposition of the two single source results. \$\endgroup\$ Dec 14, 2020 at 5:26
  • \$\begingroup\$ @LarsHankeln That is something I am not able to do. The best I have calculated is P=100*R/(3R+1)^2 . I don't want solution but verification if 10ohm is even an answer. \$\endgroup\$
    – Avezan
    Dec 14, 2020 at 5:30
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    \$\begingroup\$ @Avezan The answer is, in fact, \$R = 10\: \Omega\$. You've been told, correctly. \$\endgroup\$
    – jonk
    Dec 14, 2020 at 5:58

2 Answers 2

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Since the unknown resistance, \$R\$, and the fixed voltage supply are in series, they can be swapped without loss for analysis. The resulting schematic is as simple as the following case:

schematic

simulate this circuit – Schematic created using CircuitLab

You only have one node to solve for, making this almost trivial:

$$\frac{V}{R_1=5\:\Omega}+\frac{V}{R}=100\:\text{mS}\cdot\left(V-{-5}\:\text{V}\right)+\frac{-5\:\text{V}}{R}$$

Solve the above equation for \$V\$ and then apply it to the power equation: \$P_R=\frac{\left(V\,-\,{-5}\:\text{V}\right)^2}{R}\$. Take the differential with respect to \$R\$ and then set that equal to zero and solve. You will find the solution to be \$R=10\:\Omega\$.

In sympy:

v=Symbol('v')
r=Symbol('r')
solve(Eq(derivative(simplify((solve(Eq(v/5+v/r,.1*(v+5)-5/r),v)[0]+5)**2/r),r),0),r)[0]

10.0000000000000
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    \$\begingroup\$ @Avezan 100 mS means 100 milliSiemens. Not 100 milliseconds. It matches your schematic where you can see "0.1" on the left multiplying the voltage to set the dependent current source. \$\endgroup\$
    – jonk
    Dec 14, 2020 at 7:10
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    \$\begingroup\$ @Avezan Yes. I was able to interpret things correctly. The factor, 0.1, has units. These units are Siemens. When you multiply Siemens by Volts you get Amps. Simple dimensional analysis. So the equation is correct and the result still matches what you were told it should be. Sometimes, when someone writes "0.1" they don't give you the units, as you are supposed to "just know" what they are. I did know. Perhaps you didn't understand that. If not, this is something else you'll need to acquire in your head. Dimensional analysis is very important to learn about. \$\endgroup\$
    – jonk
    Dec 14, 2020 at 7:23
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    \$\begingroup\$ oh ok, calculation failure. 100mS is infact 0.1S, something predetermined in my head was bugging to see 10 or 0.1, I got all. Thank you very much for all your help. \$\endgroup\$
    – Avezan
    Dec 14, 2020 at 7:35
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    \$\begingroup\$ @Avezan Ah. I see. You didn't immediately convert 0.1 S to 100 mS, like I did. Got it. \$\endgroup\$
    – jonk
    Dec 14, 2020 at 7:39
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    \$\begingroup\$ @jonk Well done, nice answer (+1)! \$\endgroup\$ Dec 14, 2020 at 8:31
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First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course). Besides that the answer of @jonk is excellent.

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:

$$\text{n}\cdot\text{V}_1=\text{I}_1+\text{I}_2\tag1$$

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

$$ \begin{cases} \text{I}_1=\frac{\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_2}{\text{R}_2} \end{cases}\tag2 $$

We also know that \$\text{V}_1-\text{V}_2=\text{V}_\text{i}\$.

Now, we can solve for \$\text{V}_1\$:

$$\text{V}_1=\frac{\text{R}_1\text{V}_\text{i}}{\text{R}_1+\text{R}_2\left(1-\text{n}\text{R}_1\right)}\tag3$$

Where I used the following Mathematica-code to find \$(3)\$:

In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{n*V1 == I1 + I2, I1 == V1/R1, I2 == V2/R2, 
   V1 - V2 == Vi}, {V1, V2, I1, I2}]]

Out[1]={{V1 -> (R1 Vi)/(R1 + R2 - n R1 R2), 
  V2 -> -(((1 - n R1) R2 Vi)/(R1 + R2 - n R1 R2)), 
  I1 -> Vi/(R1 + R2 - n R1 R2), 
  I2 -> -((Vi - n R1 Vi)/(R1 + R2 - n R1 R2))}}

So, the power in the resistor \$\text{R}_1\$ is given by:

$$\text{P}_{\text{R}_1}=\frac{\text{V}_1^2}{\text{R}_1}=\left(\frac{\text{R}_1\text{V}_\text{i}}{\text{R}_1+\text{R}_2\left(1-\text{n}\text{R}_1\right)}\right)^2\cdot\frac{1}{\text{R}_1}=\text{R}_1\cdot\left(\frac{\text{V}_\text{i}}{\text{R}_1+\text{R}_2\left(1-\text{n}\text{R}_1\right)}\right)^2\tag4$$

In order to find the maximum, we can use:

$$\frac{\partial\text{P}_{\text{R}_1}}{\partial\text{R}_1}=0\space\Longleftrightarrow\space\left(\text{R}_2+\text{R}_1\left(\text{nR}_2-1\right)\right)\text{V}_\text{i}^2=0\space\Longleftrightarrow\space\text{R}_1=\frac{\text{R}_2}{1-\text{n}\text{R}_2}\tag5$$

Where I used the following Mathematica-code to find \$(5)\$:

In[2]:=Clear["Global`*"];
FullSimplify[Solve[{D[R1 *(Vi/(R1 + R2 - n R1 R2))^2, R1] == 0}, R1]]

Out[2]={{R1 -> R2/(1 - n R2)}}

Using your values, we get:

$$\text{R}_1=\frac{5}{1-\frac{1}{10}\cdot5}=10\space\Omega\tag6$$

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