Exercise 8: Answers
KVL arround abca loop:
I
1
∗
R
1
+
(
I
1
−
I
3
)
∗
R
2
+
(
I
1
−
I
2
)
∗
R
3
=
−
V
s
{\displaystyle I_{1}*R_{1}+(I_{1}-I_{3})*R_{2}+(I_{1}-I_{2})*R_{3}=-V_{s}}
Therefore
I
1
(
R
1
+
R
2
+
R
3
)
−
I
2
(
R
3
)
−
I
3
(
R
2
)
=
−
V
s
{\displaystyle I_{1}(R_{1}+R_{2}+R_{3})-I_{2}(R_{3})-I_{3}(R_{2})=-V_{s}}
KVL arround acda loop:
(
I
2
−
I
1
)
∗
R
3
+
(
I
2
−
I
3
)
∗
R
4
+
I
2
∗
R
5
=
V
s
{\displaystyle (I_{2}-I_{1})*R_{3}+(I_{2}-I_{3})*R_{4}+I_{2}*R_{5}=V_{s}}
Therefore
−
I
1
(
R
3
)
+
I
2
(
R
3
+
R
4
+
R
5
)
−
I
3
(
R
4
)
=
V
s
{\displaystyle -I_{1}(R_{3})+I_{2}(R_{3}+R_{4}+R_{5})-I_{3}(R_{4})=V_{s}}
KVL arround bcdb loop:
I
3
∗
R
6
+
(
I
3
−
I
2
)
∗
R
4
+
(
I
3
−
I
1
)
∗
R
2
=
0
{\displaystyle I_{3}*R_{6}+(I_{3}-I_{2})*R_{4}+(I_{3}-I_{1})*R_{2}=0}
Therefore
−
I
1
(
R
2
)
−
I
2
(
R
4
)
+
I
3
(
R
2
+
R
4
+
R
6
)
=
0
{\displaystyle -I_{1}(R_{2})-I_{2}(R_{4})+I_{3}(R_{2}+R_{4}+R_{6})=0}
Now we can create a matrix with the above equations as follows:
[
(
R
1
+
R
2
+
R
3
)
(
−
R
3
)
(
−
R
2
)
(
−
R
3
)
(
R
3
+
R
4
+
R
5
)
(
−
R
4
)
(
−
R
2
)
(
−
R
4
)
(
R
2
+
R
4
+
R
6
)
]
.
[
I
1
I
2
I
3
]
=
[
−
V
s
V
s
0
]
{\displaystyle {\begin{bmatrix}(R_{1}+R_{2}+R_{3})&(-R_{3})&(-R_{2})\\(-R_{3})&(R_{3}+R_{4}+R_{5})&(-R_{4})\\(-R_{2})&(-R_{4})&(R_{2}+R_{4}+R_{6})\end{bmatrix}}.{\begin{bmatrix}I_{1}\\I_{2}\\I_{3}\end{bmatrix}}={\begin{bmatrix}-V_{s}\\V_{s}\\0\end{bmatrix}}}
The following matrix is the above with values substituted:
A
.
X
→
=
Y
→
{\displaystyle A.{\vec {X}}={\vec {Y}}}
[
10220
−
10000
−
20
−
10000
30000
−
5000
−
20
−
5000
6020
]
.
[
I
1
I
2
I
3
]
=
[
−
9
9
0
]
{\displaystyle {\begin{bmatrix}10220&-10000&-20\\-10000&30000&-5000\\-20&-5000&6020\end{bmatrix}}.{\begin{bmatrix}I_{1}\\I_{2}\\I_{3}\end{bmatrix}}={\begin{bmatrix}-9\\9\\0\end{bmatrix}}}
Determinants of the General matrix, and Determinants of alterations of the general matrix as follows:
Solving determinants of:
Matrix A : General matrix A from KVL equations Matrix A1 : Genral Matrix A with Column 1 substituted by
Y
→
{\displaystyle {\vec {Y}}}
Matrix A2 : Genral Matrix A with Column 2 substituted by
Y
→
{\displaystyle {\vec {Y}}}
Matrix A3 : Genral Matrix A with Column 3 substituted by
Y
→
{\displaystyle {\vec {Y}}}
As follows:
d
e
t
A
=
9.86
×
10
11
{\displaystyle detA=9.86\times 10^{11}}
d
e
t
A
1
=
−
857700000
{\displaystyle detA1=-857700000}
d
e
t
A
2
=
11016000
{\displaystyle detA2=11016000}
d
e
t
A
3
=
6300000
{\displaystyle detA3=6300000}
I
1
;
I
2
a
n
d
I
3
{\displaystyle I_{1};I_{2}andI_{3}}
I
1
=
d
e
t
A
1
d
e
t
A
=
−
0.00086968
A
{\displaystyle I_{1}={\frac {detA1}{detA}}=-0.00086968A}
2.
I
2
=
d
e
t
A
2
d
e
t
A
=
0.00001117
A
{\displaystyle I_{2}={\frac {detA2}{detA}}=0.00001117A}
3.
I
3
=
d
e
t
A
3
d
e
t
A
=
0.000006388
A
{\displaystyle I_{3}={\frac {detA3}{detA}}=0.000006388A}
Now we can solve for the current through
R
3
{\displaystyle R_{3}}
I
R
3
=
I
1
−
I
2
=
−
0.881
m
A
{\displaystyle I_{R_{3}}=I_{1}-I_{2}=-0.881mA}
The negative sign means that
I
R
3
{\displaystyle I_{R_{3}}}
I
2
{\displaystyle I_{2}}
