Answer: will decrease by 0.43 Ohms.

Solution. Before closing the switch, the resistance of the circuit $R_{0}=R_{1}=10$ m. After closing, we get a parallel connection:

$$
R_{\mathrm{K}}=\frac{R_{1} R_{2} R_{3}}{R_{1} R_{2}+R_{1} R_{3}+R_{2} R_{3}}=\frac{1 \cdot 2 \cdot 4}{1 \cdot 2+1 \cdot 4+2 \cdot 4} \approx 0.57 \text { Ohms. }
$$

As a result, $\Delta R=R_{k}-R_{0}=-0.43$ Ohms.
5. (20 points) It is known that the resistances of the resistors in the presented circuit are \( R_{1}=1 \) Ohm, \( R_{2}=2 \) Ohms, and \( R_{3}=4 \) Ohms. Determine how and by how much the total resistance of the circuit will change when the switch \( K \) is closed.

![](https://cdn.mathpix.com/cropped/2024_05_06_0785b6f179aec901e215g-4.jpg?height=254&width=706&top_left_y=1695&top_left_x=755)