d) $\mathrm{R}_{2}=2,5 \mathrm{R}_{1}$.

Q: d) $\mathrm{R}_{2}=2,5 \mathrm{R}_{1}$.

A. True B. False $\begin{array}{l}\tan \alpha_{1}=\mathrm{R}_{1}=\frac{10}{1,25}=8 \Omega \\ \tan \alpha_{2}=\mathrm{R}_{2}=\frac{10}{0,5}=20 \Omega \\ \rightarrow \mathrm{R}_{2}=2,5 \sqrt{\mathrm{R}_{1}}\end{array}$

Question

Accepted Answer

A. True
B. False

$\begin{array}{l}\tan \alpha_{1}=\mathrm{R}_{1}=\frac{10}{1,25}=8 \Omega \ \tan \alpha_{2}=\mathrm{R}_{2}=\frac{10}{0,5}=20 \Omega \ \rightarrow \mathrm{R}_{2}=2,5 \sqrt{\mathrm{R}_{1}}\end{array}$

Đề thi cuối kì 2 (Cấu trúc mới) - Đề số 4 - MĐ 10933