Circuit Theory/2Source Excitement/Node and Mesh

The first step is to convert everything to phasors and impedances in symbolic form if possible:

${𝕍}_1=5*(\sqrt{3}+j)$

${𝕀}_1=\frac{1-j}{2*\sqrt{2}}$

$Z_{C1}=\frac{1}{j\omega C_1}=-10j$

$Z_{C2}=\frac{1}{j\omega C_2}=-5j$

$Z_L=j\omega L_1=L_2=1j$

Node Analysis

marked up for node analysis

$\frac{{𝕍}_1-{𝕍}_a}{R_1+j\omega L_1}+{𝕀}_1-\frac{{𝕍}_a-{𝕍}_b}{\frac{1}{j\omega C_1}}=0$

$\frac{{𝕍}_a-{𝕍}_b}{\frac{1}{j\omega C_1}}-\frac{{𝕍}_b}{R_3}-\frac{{𝕍}_b-{𝕍}_c}{\frac{1}{j\omega C_2}}=0$

$\frac{{𝕍}_b-{𝕍}_c}{\frac{1}{j\omega C_2}}-{𝕀}_1-\frac{{𝕍}_c}{j\omega L_2}=0$

Results using matlab:

${𝕍}_a=-6.5-7i\Rightarrow V_a(t)=9.55cos(1000t-2.32)$

${𝕍}_b=-3.08-3.31i\Rightarrow V_b(t)=4.52cos(1000t-2.32)$

${𝕍}_c=0.327+0.385i\Rightarrow V_c(t)=0.506cos(1000t+0.866)$

Mesh Analysis

marked up for mesh analysis

${𝕀}_1{R}_2+\frac{{𝕀}_1+{𝕀}_2}{j\omega C_1}+\frac{{𝕀}_1+{𝕀}_3}{j\omega C_2}-{𝕍}_{Is}=0$

${𝕍}_1+{𝕀}_2{R}_1+\frac{{𝕀}_2+{𝕀}_1}{j\omega C_1}+({𝕀}_2-{𝕀}_3)R_3+{𝕀}_{2}j\omega L_1=0$

$\frac{{𝕀}_3+{𝕀}_1}{j\omega C_2}+{𝕀}_{3}j\omega L_2+({𝕀}_3-{𝕀}_2)R_3=0$

Results using Matlab:

${𝕀}_2=-0.239+0.234i\Rightarrow I_2(t)=0.334cos(1000t+2.37)$

${𝕀}_3=-0.238+0.235i\Rightarrow I_3(t)=0.334cos(1000t+2.36)$

${𝕍}_{Is}=175-179i\Rightarrow V_{Is}(t)=250cos(1000t-0.795)$