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This problem asks me to find Rx, the electrical power of each element and the algebraic sum of the electrical powers of all elements. How to apply Kirchhoff's and Ohm's law correctly in this problem?

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Up to this time, I have have found Ux=200V and I tried finding Rx by saying that Ix=V/(50+Rx)=>6+2=(1200+10*200)/(Rx+50)=>Rx=350Ω which I think is incorrect.

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Well, the circuit we want to analyze is given by:

schematic

simulate this circuit – Schematic created using CircuitLab

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

$$ \text{I}_1=\text{I}_2+\text{I}_3\tag1 $$

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

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\displaystyle\text{V}_1-0}{\displaystyle\text{R}_1}\\ \\ \text{I}_2&=\frac{\displaystyle\text{V}_2-\text{V}_1}{\displaystyle\text{R}_2}\\ \\ \text{I}_2&=\frac{\displaystyle0-\text{V}_4}{\displaystyle\text{R}_3} \end{alignat*} \end{cases}\tag2 $$

We also know that: \$\displaystyle\text{V}_2-\text{V}_3=\text{n}\cdot\text{V}_1\$ and \$\displaystyle\text{V}_3-\text{V}_4=\text{V}_\text{i}\$.

Filling in what we know, gives:

$$ \begin{cases} \begin{alignat*}{1} -2&=\text{I}_2+6\\ \\ -2&=\frac{\displaystyle\text{V}_1-0}{\displaystyle100}\\ \\ \text{I}_2&=\frac{\displaystyle\text{V}_2-\text{V}_1}{\displaystyle50}\\ \\ \text{I}_2&=\frac{\displaystyle0-\text{V}_4}{\displaystyle\text{R}_3}\\ \\ \text{V}_2-\text{V}_3&=10\cdot\text{V}_1\\ \\ \text{V}_3-\text{V}_4&=1200 \end{alignat*} \end{cases}\tag3 $$

Solving \$\displaystyle\left(3\right)\$, gives:

$$\text{R}_3=25\space\Omega,\text{I}_2=-8\space\text{A},\text{V}_1=-200\space\text{V},\text{V}_2=-600\space\text{V},\text{V}_3=1400\space\text{V},\text{V}_4=200\space\text{V}\tag4$$

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