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