An example of Kirchhoff rules.
--> I1 I2 -->
a R1 b R2 c
+---/\/\/---+---/\/\/---+
| | |
| I3 < |
___ + | > R5 ___ +
- E1 v < - E2
| < |
| | |
+---/\/\/---+---/\/\/---+
d R3 e R4 f
b & e:
$$\tag{1}
I_1 - I_2 - I_3 = 0
$$
a-b-e-d-a:
$$\tag{3}
\varepsilon_1 - I_1 R_1 - I_3 R_5 - I_1 R_3 = 0
$$
c-b-e-f-c:
$$\tag{4}
\varepsilon_2 + I_2 R_2 - I_3 R_5 + I_2 R_4 = 0
$$
a-b-c-f-e-d-a | (3) - (4):
$$\tag{5}
\begin{array}{rcl}
\varepsilon_1 - I_1 R_1 - I_2 R_2 &&\newline
-\varepsilon_2 - I_2 R_4 - I_1 R_3 & = & 0
\end{array}
$$
(1) → (2):
$$\tag{6}
\begin{array}{rcl}
\varepsilon_1 \newline
-(R_1+R_3+R_5) I_1 \newline
+R_5 I_2 & = & 0
\end{array}
$$
(1) → (3):
$$\tag{7}
\begin{array}{rcl}
\varepsilon_2 \newline
-R_5 I_1 \newline
+(R_2+R_4+R_5) I_2 & = & 0
\end{array}
$$
$R_5$ (6) - $(R_1+R_3+R_5)$ (7):
$$\tag{8}
\begin{array}{c}
R_5 \varepsilon_1 - (R_1+R_3+R_5) \varepsilon_2 \newline
-(R_1+R_3+R_5)(R_2+R_4+R_5) I_2 \newline
+R_5^2 I_2 = 0
\end{array}
$$
$$\tag{9}
\begin{array}{c}
R_5 \varepsilon_1 - (R_1+R_3+R_5) \varepsilon_2 \newline
-[(R_1+R_3+R_4+R_4)R_5 \newline
+(R_1+R_3)(R_2+R_4)] I_2 = 0
\end{array}
$$
$$\tag{10}
I_2 = \frac{
R_5 \varepsilon_1 - (R_1+R_3+R_5) \varepsilon_2
}{
(R_1+R_3+R_4+R_4)R_5 + (R_1+R_3)(R_2+R_4)
}
$$
•