# Problem № 7 (10 points)

In the electrical circuit shown in the diagram, the resistances of the resistors are $R_{1}=10$ Ohms and $R_{2}=20$ Ohms. A current source is connected to points A and B in the circuit. When the polarity of its connection is reversed, the ammeter readings change by one and a half times. Determine the internal resistance of the source. Consider the ammeter to be ideal. Assume that the resistance of the diodes in the forward direction is negligibly small, and in the reverse direction, it is infinitely large.

![](https://cdn.mathpix.com/cropped/2024_05_06_7ac9c551b647ccda756fg-3.jpg?height=528&width=859&top_left_y=798&top_left_x=610)

## Answer: 10 Ohms

#
# Solution and Evaluation Criteria:

When the positive terminal of the power supply is connected to point $A$, the current flows only through resistor $R_{2}$, and in this case: $I_{1}=\frac{\varepsilon}{R_{2}+r}$.

## (3 points)

When the polarity is reversed, the current flows only through resistance $R_{1}$, and:

$I_{2}=\frac{\varepsilon}{R_{1}+r}$

According to the condition: $I_{2}=1.5 I_{1}$

We get:

$$
\begin{aligned}
& 1.5 I_{1}\left(R_{1}+r\right)=I_{1}\left(R_{2}+r\right) \\
& 1.5(10+r)=20+r \\
& r=10 \Omega .
\end{aligned}
$$