Câu hỏi 50

+ Ban đầu: H = 80%

\(\Delta {{P}_{1}}=\frac{P_{1}^{2}R}{U_{1}^{2}\text{co}{{\text{s}}^{2}}\varphi }=0,2{{P}_{1}}\Rightarrow {{P}_{1}}=\frac{0,2U_{1}^{2}}{R}\)

\({{P}_{1}}'=0,8{{P}_{1}}=\frac{0,8.0,2.U_{1}^{2}}{R}\)

+ Lúc sau: Hiệu suất là H

\(\Delta {{P}_{2}}=\frac{P_{2}^{2}R}{U_{2}^{2}\text{co}{{\text{s}}^{2}}\varphi }=(1-H){{P}_{2}}\Rightarrow {{P}_{2}}=\frac{(1-H)U_{2}^{2}}{R}\)

\({{P}_{2}}'=H{{P}_{1}}=\frac{H(1-H).U_{2}^{2}}{R}\)

Vì P₁’ = P₂’ => 0,8.0,2.U₁² = H(1-H)U₂² => 0,8.0,2 = 4H(1-H) => H = 0,958 = 95,8%

Chọn D