Answer: $24$ mA

Solution. As a result of analyzing the proposed electrical circuit, it can be concluded that: $I_{2}=I_{1}=2$ mA.

\[
\begin{aligned}
& I_{3}=2 I_{1}=4 \mathrm{mA} \\
& I_{5}=I_{3}+I_{1}=6 \mathrm{mA} \\
& I_{4}=\frac{5}{3} I_{5}=10 \mathrm{mA}
\end{aligned}
\]

The sum of the readings of all ammeters: $I=I_{1}+I_{2}+I_{3}+I_{4}+I_{5}=24$ mA. (2 points)
5. (10 points) Five identical non-ideal ammeters are connected as shown in the figure. Ideal power supply is connected to points $A$ and $B$. Determine the sum of the readings of all ammeters, given that the reading of the first ammeter $I_{1}=2 \mathrm{MA}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9c0cded10981f14efe49g-03.jpg?height=774&width=1108&top_left_y=458&top_left_x=494)