[<< wikibooks] 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
∗
(

3

+
j
)

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

I

1

=

1
−
j

2
∗

2

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

Z

C
1

=

1

j
ω

C

1

=
−
10
j

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

Z

C
2

=

1

j
ω

C

2

=
−
5
j

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

Z

L

=
j
ω

L

1

=

L

2

=
1
j

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

== Node Analysis ==

V

1

−

V

a

R

1

+
j
ω

L

1

+

I

1

−

V

a

−

V

b

1

j
ω

C

1

=
0

{\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}

V

a

−

V

b

1

j
ω

C

1

−

V

b

R

3

−

V

b

−

V

c

1

j
ω

C

2

=
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}

V

b

−

V

c

1

j
ω

C

2

−

I

1

−

V

c

j
ω

L

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:

V

a

=
−
6.5
−
7
i
⇒

V

a

(
t
)
=
9.55
c
o
s
(
1000
t
−
2.32
)

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

V

b

=
−
3.08
−
3.31
i
⇒

V

b

(
t
)
=
4.52
c
o
s
(
1000
t
−
2.32
)

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

V

c

=
0.327
+
0.385
i
⇒

V

c

(
t
)
=
0.506
c
o
s
(
1000
t
+
0.866
)

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

== Mesh Analysis ==

I

1

R

2

+

I

1

+

I

2

j
ω

C

1

+

I

1

+

I

3

j
ω

C

2

−

V

I
s

=
0

{\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}

V

1

+

I

2

R

1

+

I

2

+

I

1

j
ω

C

1

+
(

I

2

−

I

3

)

R

3

+

I

2

j
ω

L

1

=
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}

I

3

+

I

1

j
ω

C

2

+

I

3

j
ω

L

2

+
(

I

3

−

I

2

)

R

3

=
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:

I

2

=
−
0.239
+
0.234
i
⇒

I

2

(
t
)
=
0.334
c
o
s
(
1000
t
+
2.37
)

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

I

3

=
−
0.238
+
0.235
i
⇒

I

3

(
t
)
=
0.334
c
o
s
(
1000
t
+
2.36
)

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

V

I
s

=
175
−
179
i
⇒

V

I
s

(
t
)
=
250
c
o
s
(
1000
t
−
0.795
)

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