5. (20 points) It is known that the resistances of the resistors in the presented circuit are \( R_{1}=4 \) Ohms, \( R_{2}=8 \) Ohms, and \( R_{3}=16 \) 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-2.jpg?height=248&width=694&top_left_y=1669&top_left_x=767)
Answer: will decrease by 1.7 Ohms.

Solution. Before closing the switch, the resistance of the circuit $R_{0}=R_{1}=4$ Ohms. After closing, we get a parallel connection:

$$
R_{\mathrm{K}}=\frac{R_{1} R_{2} R_{3}}{R_{1} R_{2}+R_{1} R_{3}+R_{2} R_{3}}=\frac{4 \cdot 8 \cdot 16}{4 \cdot 8+4 \cdot 16+8 \cdot 16} \approx 2.30 \mathrm{Ohms}.
$$

As a result, $\Delta R=R_{k}-R_{0}=-1.7$ Ohms.