Circuit Theory/2Source Excitement/Node and Mesh

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

${\displaystyle \mathbb {V} _{1}=5*({\sqrt {3}}+j)}$

${\displaystyle \mathbb {I} _{1}={\frac {1-j}{2*{\sqrt {2}}}}}$

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

${\displaystyle Z_{C2}={\frac {1}{j\omega C_{2}}}=-5j}$

${\displaystyle Z_{L}=j\omega L_{1}=L_{2}=1j}$

${\displaystyle {\frac {\mathbb {V} _{1}-\mathbb {V} _{a}}{R_{1}+j\omega L_{1}}}+\mathbb {I} _{1}-{\frac {\mathbb {V} _{a}-\mathbb {V} _{b}}{\frac {1}{j\omega C_{1}}}}=0}$

${\displaystyle {\frac {\mathbb {V} _{a}-\mathbb {V} _{b}}{\frac {1}{j\omega C_{1}}}}-{\frac {\mathbb {V} _{b}}{R_{3}}}-{\frac {\mathbb {V} _{b}-\mathbb {V} _{c}}{\frac {1}{j\omega C_{2}}}}=0}$

${\displaystyle {\frac {\mathbb {V} _{b}-\mathbb {V} _{c}}{\frac {1}{j\omega C_{2}}}}-\mathbb {I} _{1}-{\frac {\mathbb {V} _{c}}{j\omega L_{2}}}=0}$

Results using matlab:

${\displaystyle \mathbb {V} _{a}=-6.5-7i\Rightarrow V_{a}(t)=9.55cos(1000t-2.32)}$

${\displaystyle \mathbb {V} _{b}=-3.08-3.31i\Rightarrow V_{b}(t)=4.52cos(1000t-2.32)}$

${\displaystyle \mathbb {V} _{c}=0.327+0.385i\Rightarrow V_{c}(t)=0.506cos(1000t+0.866)}$

${\displaystyle \mathbb {I} _{1}R_{2}+{\frac {\mathbb {I} _{1}+\mathbb {I} _{2}}{j\omega C_{1}}}+{\frac {\mathbb {I} _{1}+\mathbb {I} _{3}}{j\omega C_{2}}}-\mathbb {V} _{Is}=0}$

${\displaystyle \mathbb {V} _{1}+\mathbb {I} _{2}R_{1}+{\frac {\mathbb {I} _{2}+\mathbb {I} _{1}}{j\omega C_{1}}}+(\mathbb {I} _{2}-\mathbb {I} _{3})R_{3}+\mathbb {I} _{2}j\omega L_{1}=0}$

${\displaystyle {\frac {\mathbb {I} _{3}+\mathbb {I} _{1}}{j\omega C_{2}}}+\mathbb {I} _{3}j\omega L_{2}+(\mathbb {I} _{3}-\mathbb {I} _{2})R_{3}=0}$

Results using Matlab:

${\displaystyle \mathbb {I} _{2}=-0.239+0.234i\Rightarrow I_{2}(t)=0.334cos(1000t+2.37)}$

${\displaystyle \mathbb {I} _{3}=-0.238+0.235i\Rightarrow I_{3}(t)=0.334cos(1000t+2.36)}$

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