Answer: 2.08 Ω

Solution. In the given situation, the following equivalent circuit is obtained (5 points):

![](https://cdn.mathpix.com/cropped/2024_05_06_564114d217ece4a63410g-05.jpg?height=326&width=791&top_left_y=2093&top_left_x=632)

Its total resistance: $R=\frac{R_{1} R_{3}}{R_{1}+R_{3}}+\frac{R_{2} R_{4}}{R_{2}+R_{4}}=\frac{1 \cdot 3}{1+3}+\frac{2 \cdot 4}{2+4}=\frac{25}{12} \approx 2.08$ Ω (5 points).
7. In the electrical circuit depicted in the figure, the resistances of the resistors are $R_{1}=1 \Omega$, $R_{2}=2 \Omega$, $R_{3}=3 \Omega$, and $R_{4}=4 \Omega$. Assume that the resistances of all diodes in the forward direction are negligibly small, and in the reverse direction are infinite. Determine the resistance of the entire circuit between points $A$ and $B$ in the situation where the positive pole of the current source is connected to point $A$, and the negative pole to point $B$. The resistance of the connecting wires is negligibly small. (10 points)

![](https://cdn.mathpix.com/cropped/2024_05_06_564114d217ece4a63410g-05.jpg?height=474&width=759&top_left_y=1334&top_left_x=657)