4. The electrical circuit shown in the figure contains a very large number of identical resistors and identical voltmeters. It is known that the readings of voltmeters $V_{1}$ and $V_{2}$ are $U=4$ V and $U^{\prime}=6$ V (although it is unknown which reading corresponds to which voltmeter). Find the readings of voltmeters $V_{3}$ and $V_{4}$, as well as the sum of the readings of all voltmeters.

![](https://cdn.mathpix.com/cropped/2024_05_06_2fcbf33eea2a4c6aa801g-1.jpg?height=266&width=691&top_left_y=1295&top_left_x=1248)
4. Since the current flowing through the voltmeter $V_{1}$ is greater than the current flowing through the voltmeter $V_{2}$, the resistances of the voltmeters are the same, and each voltmeter shows the voltage across it, the value $U^{\prime}=6$ V is the reading of the voltmeter $V_{1}$, $U=4$ V is the reading of the voltmeter $V_{2}$, and the ratio of the currents flowing through the voltmeters $V_{1}$ and $V_{2}$ is

$$
\frac{I_{1}}{I_{2}}=\frac{U^{\prime}}{U}=\frac{3}{2} \quad \Rightarrow \quad I_{2}=\frac{2}{3} I_{1}
$$

Since the current flowing through the voltmeter $V_{1}$ is divided into the currents flowing through $V_{2}$ and $V_{3}$, the current flowing through the voltmeter $V_{3}$ is

$$
I_{3}=\frac{1}{3} I_{1}
$$

Therefore, the resistance of the circuit segment shown in the figure on the right is twice the resistance of the voltmeter $R_{x}=2 R$, where $R$ is the resistance of the voltmeter, and $R_{x}$ is the resistance of the circuit segment shown in the figure. On the other hand, if one link is removed from the infinite circuit, its resistance does not change.

![](https://cdn.mathpix.com/cropped/2024_05_06_69aec43348e730276044g-2.jpg?height=257&width=488&top_left_y=1479&top_left_x=1475)

Therefore, the resistance of the circuit shown in the figure is also equal to $R_{x}$. Hence,

$$
R_{x}=r+R+\frac{R R_{x}}{R+R_{x}}
$$

where $r$ is the resistance of the resistor. Considering that $R_{x}=2 R$, we find

$$
r=\frac{R}{3}
$$

Therefore, the voltage across the resistor in the first link of the considered circuit $U_{r}=2 \mathrm{~V}$, and the voltage created by the source across the entire circuit is

$$
U=U_{r}+U^{\prime}+U=12 \mathrm{~V} .
$$

The voltage applied to the second link of our circuit is three times less, $U=4$ V. This means that the readings of the voltmeters in the second link of the circuit $V_{3}$ and $V_{4}$ will be three times less than the readings of the voltmeters in the first link

$$
U_{3}=2 \mathrm{~V}, U_{4}=\frac{4}{3} \mathrm{~V}
$$

That is, their sum will be equal to

$$
U_{3}+U_{4}=\frac{10}{3}=\frac{1}{3}\left(U_{1}+U_{2}\right)
$$

Similarly, the sum of the readings of the voltmeters $V_{5}$ and $V_{6}$ will be one third of the sum of the readings of the voltmeters $V_{3}$ and $V_{4}$

$$
U_{5}+U_{6}=\frac{1}{3}\left(U_{3}+U_{4}\right)=\frac{1}{9}\left(U_{1}+U_{2}\right)
$$

From this, we conclude that the sum of the readings of all voltmeters will be equal to

$$
\sum U=U_{1}+U_{2}+U_{3}+U_{4}+\ldots=\left(U_{1}+U_{2}\right)\left(1+\frac{1}{3}+\frac{1}{9}+\ldots\right)
$$

Finding the sum of the geometric progression in parentheses, we get

$$
\sum U=\frac{3}{2}\left(U+U^{\prime}\right)=15 \mathrm{~V}
$$