Find v1=? and v2=? for the circuit

Question

This is my solution,but i have a problem with doing a node analysis for node 2 and 3.In my textbook the solution is like this: (2+i1+i1)V1 -2VO -i1V2=0 and(1-i2+i1)V2-i1V1=5

Answer 1

The solution in your text book is correct:

The solution would be \$V_1 = 2 - j\$ and \$V_2 = 2+4\cdot j\$.

Answer 2

Because you already have the answer I will present my answer using Mathematica. This way you can solve your circuit more generally.

Well, let's solve this mathematically. We have the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

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

$$ \begin{cases} \text{I}_1=\text{I}_2+\text{I}_7\\ \\ \text{I}_7=\text{I}_3+\text{I}_4\\ \\ \text{I}_8=\text{I}_3+\text{I}_4\\ \\ \text{I}_8=\text{I}_5+\text{I}_9\\ \\ \text{I}_6=\text{I}_9+\text{I}_\text{b}\\ \\ \text{I}_6=\text{I}_{10}+\text{I}_\text{b}\\ \\ \text{I}_{11}=\text{I}_5+\text{I}_{10}\\ \\ \text{I}_1=\text{I}_2+\text{I}_{11} \end{cases}\tag1 $$

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

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{a}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_1-\text{V}_2}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_2}{\text{R}_5}\\ \\ \text{I}_6=\frac{\text{V}_2}{\text{R}_6} \end{cases}\tag2 $$

Now, we can use Mathematica to write a code that solves all the unknowns:

Va =;
Ib =;
R1 =;
R2 =;
R3 =;
R4 =;
R5 =;
R6 =;
FullSimplify[
 Solve[{I1 == I2 + I7, I7 == I3 + I4, I8 == I3 + I4, I8 == I5 + I9, 
   I6 == I9 + Ib, I6 == I10 + Ib, I11 == I5 + I10, I1 == I2 + I11, 
   I1 == (Va - V1)/R1, I2 == (V1)/R2, I3 == (V1 - V2)/R3, 
   I4 == (V1 - V2)/R4, I5 == (V2)/R5, I6 == (V2)/R6}, {I1, I2, I3, I4,
    I5, I6, I7, I8, I9, I10, I11, V1, V2}]]

---

Now, applying this to your circuit we need to use (we use the 'complex' s-domain where I used Laplace transform):

• $$\text{R}_2=\frac{1}{\text{sC}_1}\tag3$$
• $$\text{R}_3=\text{sL}_1\tag4$$
• $$\text{R}_4=\frac{1}{\text{sC}_2}\tag5$$
• $$\text{R}_5=\text{sL}_2\tag6$$

So, for the code we get:

Va = LaplaceTransform[u*Cos[\[Omega]1*t], t, s];
Ib = LaplaceTransform[i*Cos[\[Omega]2*t], t, s];
u =;
i =;
\[Omega]1 =;
\[Omega]2 =;
R1 =;
R2 = 1/(s*C1);
R3 = s*L1;
R4 = 1/(s*C2);
R5 = s*L2;
R6 =;
C1 =;
C2 =;
L1 =;
L2 =;
FullSimplify[
 Solve[{I1 == I2 + I7, I7 == I3 + I4, I8 == I3 + I4, I8 == I5 + I9, 
   I6 == I9 + Ib, I6 == I10 + Ib, I11 == I5 + I10, I1 == I2 + I11, 
   I1 == (Va - V1)/R1, I2 == (V1)/R2, I3 == (V1 - V2)/R3, 
   I4 == (V1 - V2)/R4, I5 == (V2)/R5, I6 == (V2)/R6}, {I1, I2, I3, I4,
    I5, I6, I7, I8, I9, I10, I11, V1, V2}]]

Solving this in your circuit, gives:

In[1]:=Va = LaplaceTransform[u*Cos[\[Omega]1*t], t, s];
Ib = LaplaceTransform[i*Cos[\[Omega]2*t], t, s];
u = 5;
i = 5;
\[Omega]1 = 2;
\[Omega]2 = 2;
R1 = 1/2;
R2 = 1/(s*C1);
R3 = s*L1;
R4 = 1/(s*C2);
R5 = s*L2;
R6 = 1;
C1 = 1/2;
C2 = 1;
L1 = 1/2;
L2 = 1/4;
FullSimplify[
 Solve[{I1 == I2 + I7, I7 == I3 + I4, I8 == I3 + I4, I8 == I5 + I9, 
   I6 == I9 + Ib, I6 == I10 + Ib, I11 == I5 + I10, I1 == I2 + I11, 
   I1 == (Va - V1)/R1, I2 == (V1)/R2, I3 == (V1 - V2)/R3, 
   I4 == (V1 - V2)/R4, I5 == (V2)/R5, I6 == (V2)/R6}, {I1, I2, I3, I4,
    I5, I6, I7, I8, I9, I10, I11, V1, V2}]]

Out[1]={{I1 -> (10 s (16 + s^2 (14 + s + s^2)))/((1 + s) (4 + s) (4 + 
      s^2) (4 + s (2 + s))), 
  I2 -> (5 s^3 (14 + s (2 + 3 s)))/((1 + s) (4 + s) (4 + s^2) (4 + 
      s (2 + s))), 
  I3 -> -((10 (-4 + s) s)/((1 + s) (4 + s^2) (4 + s (2 + s)))), 
  I4 -> -((5 (-4 + s) s^3)/((1 + s) (4 + s^2) (4 + s (2 + s)))), 
  I5 -> (20 s (12 + s (4 + 7 s)))/((1 + s) (4 + s) (4 + s^2) (4 + 
      s (2 + s))), 
  I6 -> (5 s^2 (12 + s (4 + 7 s)))/((1 + s) (4 + s) (4 + s^2) (4 + 
      s (2 + s))), 
  I7 -> -((5 (-4 + s) s (2 + s^2))/((1 + s) (4 + s^2) (4 + 
       s (2 + s)))), 
  I8 -> -((5 (-4 + s) s (2 + s^2))/((1 + s) (4 + s^2) (4 + 
       s (2 + s)))), 
  I9 -> -((5 s (16 + s (16 + 14 s + s^3)))/((1 + s) (4 + s) (4 + 
       s^2) (4 + s (2 + s)))), 
  I10 -> -((
    5 s (16 + s (16 + 14 s + s^3)))/((1 + s) (4 + s) (4 + s^2) (4 + 
       s (2 + s)))), 
  I11 -> -((
    5 (-4 + s) s (2 + s^2))/((1 + s) (4 + s^2) (4 + s (2 + s)))), 
  V1 -> (10 s^2 (14 + s (2 + 3 s)))/((1 + s) (4 + s) (4 + s^2) (4 + 
      s (2 + s))), 
  V2 -> (5 s^2 (12 + s (4 + 7 s)))/((1 + s) (4 + s) (4 + s^2) (4 + 
      s (2 + s)))}}