Contradiction while solving circuit problem!

Translation: Find the supplied power from the 4V Source using meshes method

I proposed five equations related with the circuit but I only find contradiction cause I find Rv value fixed. I suspect there should be voltage drops in current sources but I am not sure I think I am doing something conceptually wrong.

Here my calcuations

I skipped some steps but I hope you can follow me.

I am currently getting I2=I3 = 0 which makes the first two equations to fix Rv. By my understanding it is a contradiction.

Note: .5Nk is a name for the resistor value (proportional to N), I have just called it Rv.

Answer

Let's set up the mesh equations as you have the schematic drawn up:

$$\begin{align*} 0\:\text{V}-200\cdot I_3-600\cdot I_3-400\cdot\left(I_3-I_2\right)-V_{5\:\text{mA}}&= 0\:\text{V}\\\\ 0\:\text{V}-R_V\cdot\left(I_2-I_1\right)-400\cdot\left(I_2-I_3\right)-V_{V_2\over 400}&=0\:\text{V}\\\\ 0\:\text{V}+4\:\text{V}-500\cdot I_1+V_{5\:\text{mA}}-R_V\cdot\left(I_1-I_2\right)&=0\:\text{V}\\\\ I_1-I_3&=5\:\text{mA}\\\\ I_2=\frac{V_2=200\cdot I_3}{400}&=\frac{I_3}{2} \end{align*}$$

This provides 5 equations and five unknowns: $I_1$, $I_2$, $I_3$, $V_{5\:\text{mA}}$, $V_{V_2\over 400}$.

Note that this includes the fact that there are, in fact, voltage drops across your current sources. Those are just two more variables, as shown.

Solving this yields 5 functions which depend upon $R_V$.

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I can't tell you if $I_2=I_3=0\:\text{A}$ without knowing more about $R_V$. And I don't understand $.5\:N\:k$. (Does it mean $500\cdot N$?) So I'm stuck at this point, if you are looking for numerical results that aren't functions of $R_V$ (or N.)

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(You can get $I_2=I_3=0\:\text{A}$ if and only if $R_V=300\:\Omega$. For obvious reasons.)