Hello,

In this circuit, i need to find the node voltages for v1 and v2.

I have for V1: \(0.1v_{1}+J0.05v_{1} = 0\)

and for V2: \(0.067V_{2}-J0.2V_{2} = 0\)

And V1 = V2 + 10

Substituting for V1:

\(0.1\cdot(v_{2}+10) +J0.05v_{2}+J0.5\)

Am i on the wrong track?

 

The question is presented to use loop analysis. Here's a starter:

Loop 1: \(I_1 (10 \Omega) + (I_1 - I_2)(j20 \Omega) = 0\)
Loop 2: \((I_2 - I_1)(j20 \Omega) + (I_2 - I_3)(15 \Omega) = 10V\)
Loop 3: \((I_3 - I_2)(15 \Omega) + I_3(-j5 \Omega) = 0\)

Note: the voltage source symbol is that of a DC source, but the question suggests AC.

 

The question is presented to use loop analysis. Here's a starter:

Loop 1: \(I_1 (10 \Omega) + (I_1 - I_2)(j20 \Omega) = 0\)
Loop 2: \((I_2 - I_1)(j20 \Omega) + (I_2 - I_3)(15 \Omega) = 10V\)
Loop 3: \((I_3 - I_2)(15 \Omega) + I_3(-j5 \Omega) = 0\)

Note: the voltage source symbol is that of a DC source, but the question suggests AC.

Hm, can you really do mesh analysis like that when you have a supernode?

 

Hm, can you really do mesh analysis like that when you have a supernode?

No -- supernodes are used in nodal analysis. Transforms to go from one form of analysis to another exist (Thevenin, Norton), but usually only complicate matters. I'm not sure what you're calling a "supernode", here, but I know it as the combination of 2 or more nodes.

 

The question is presented to use loop analysis. Here's a starter:

Loop 1: \(I_1 (10 \Omega) + (I_1 - I_2)(j20 \Omega) = 0\)
Loop 2: \((I_2 - I_1)(j20 \Omega) + (I_2 - I_3)(15 \Omega) = 10V\)
Loop 3: \((I_3 - I_2)(15 \Omega) + I_3(-j5 \Omega) = 0\)

Note: the voltage source symbol is that of a DC source, but the question suggests AC.

\(\left[ {\begin{array}{cc}
(10+J20) & -J20 & 0 \\
-J20 & (15+J20) &-15 \\
0 & -15 &(15-J5)\end{array}}\right] \left[ {\begin{array}{cc}I_{1}\\
I_{2}\\
I_{3}\end{array}}\right]\)

Hm, it's easy to see that the middle row becomes zero

Can you just add the rows
So that you get:
\(10I_{1} -J5I_{3} = 10\)

 

The question is presented to use loop analysis. Here's a starter:

Loop 1: \(I_1 (10 \Omega) + (I_1 - I_2)(j20 \Omega) = 0\)
Loop 2: \((I_2 - I_1)(j20 \Omega) + (I_2 - I_3)(15 \Omega) = 10V\)
Loop 3: \((I_3 - I_2)(15 \Omega) + I_3(-j5 \Omega) = 0\)

Note: the voltage source symbol is that of a DC source, but the question suggests AC.

The red circled arrows do suggest a mesh analysis, but it would be just as reasonable to infer from the marked node voltages V1 and V2 that a node analysis is wanted.

Your Loop 2 equation has the wrong sign on the right hand side; it should be -10V.

Hm, it's easy to see that the middle row becomes zero

Can you just add the rows
So that you get:
\(10I_{1} -J5I_{3} = 10\)

This gives you the relationship between I1 and I3, but you need V1 and V2.

The nodal equations would be:

\(\left[ {\begin{array}{cc}
\frac{1}{10}+\frac{1}{j20} & \frac{1}{15}+\frac{1}{-j5} \\ 1 & -1 \end{array}}\right] \left[ {\begin{array}{cc}V_{1}\\
V_{2} \end{array}}\right]=\left[ \begin{array}{cc}0\\10\\ \end{array}\right]\)

The second equation (row) is the constraint equation to deal with the fact that the two nodes are connected by a voltage source (they are a supernode).