Between points $A$ and $B$, 3 branches are connected. In the first branch, there is a 2 Volt electromotive force element and a 2 Ohm resistance connected in series, in the second branch, there is a 2 Volt element and a $1 \Omega$ resistance, and the third branch is formed by a conductor with a resistance of $1 \Omega$. What is the current in each branch and what is the voltage between $A$ and $B$?
Consider the following circuit diagram: the electromotive forces of the elements are $E_{1}$ and $E_{2}$, the resistances are $r_{1}, r_{2}$, $r_{3}$, and the currents in the respective branches are $i_{1}, i_{2}, i_{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_3b5f8c02cc230373b8f9g-1.jpg?height=127&width=236&top_left_y=240&top_left_x=934)

Based on Kirchhoff's laws, we can write the equations for the closed loops $\left(E_{1} A r_{3} B r_{1}\right)$ and $\left(E_{2} A r_{3} B r_{2}\right)$ as follows:

$$
\begin{aligned}
& E_{1}=i_{1} r_{1}+i_{3} r_{3} \\
& E_{2}=i_{2} r_{2}+i_{3} r_{3}
\end{aligned}
$$

Furthermore, for the current flowing into and out of point $A$:

$$
i_{3}=i_{1}+i_{2}
$$

Substituting the numerical values into these equations:

$$
2=2 i_{1}+i_{3} ; \quad 2=i_{2}+i_{3} ; \quad i_{3}=i_{1}+i_{2}
$$

Using the third equation in the first two:

$$
2=3 i_{1}+i_{2} ; \quad 2=i_{1}+2 i_{2}
$$

Subtracting the appropriate sides:

$$
0=-2 i_{1}+i_{2} ; \quad i_{2}=2 i_{1}
$$

From the first equation:

$$
\begin{aligned}
5 i_{1}=2, \quad i_{1} & =0.4 \text { Ampere } \\
i_{2} & =0.8 \quad " \\
i_{3} & =1.2 \quad "
\end{aligned}
$$

The voltage between points $A$ and $B$ (denoted by $E$) can be calculated in the third branch using Ohm's law:

$$
r_{3}=\frac{E}{i_{3}} ; \quad E=i_{3} r_{3}=1.2 \cdot 1=1.2 \text { Volt. }
$$