The elements of a settlement made of Bunsen cells are connected in series. The electromotive force of each element is 1.8 Volts, and the internal resistance is 0.2 ohms. The two ends are connected in parallel with two wires, $A$ and $B$. The resistance of $A$ is 6 ohms, and that of $B$ is 3 ohms. How many amperes is the total current intensity and how many amperes flow in each of the wires.
The electromotive force of the circuit is $1.8 \cdot 6 = 10.8 \mathrm{~V}$.

The internal resistance of the cell: $0.2 \cdot 6 = 1.2 \Omega$.

The external resistance of the circuit: $\frac{6 \cdot 3}{6+3} = \frac{18}{9} = 2 \Omega$.

Therefore, the total resistance is: $1.2 + 2 = 3.2 \Omega$, and according to Ohm's law, the main current intensity is:

$$
I = \frac{10.8}{3.2} \cdot \frac{\mathrm{V}}{\Omega} = 3.38 \mathrm{~A}
$$

This main current is divided between the $A$ and $B$ wires in the inverse ratio of the resistances

$$
\frac{i_{1}}{i_{2}} = \frac{3}{6} = \frac{1}{2} ; \quad i_{1} = \frac{1}{2} i_{2}
$$

furthermore

$$
i_{1} + i_{2} = 3.38 \mathrm{~A}
$$

thus

$$
\begin{gathered}
\frac{1}{2} i_{2} + i_{2} = 3.38 \mathrm{~A} \\
\frac{3}{2} i_{2} = 3.38 \mathrm{~A} \\
i_{2} = \frac{2}{3} \cdot 3.38 \mathrm{~A} = 2.25 \mathrm{~A} \\
i_{1} = \frac{1}{3} \cdot 3.38 \mathrm{~A} = 1.13 \mathrm{~A}
\end{gathered}
$$